W11-W12. Conic Sections, Quadratic Forms, Tangent Lines

Author

Salman Ahmadi-Asl

Published

November 13, 2025

1. Summary

1.1 Conic Sections Overview

Conic sections are curves obtained by intersecting a plane with a double cone. The three main types are ellipses, hyperbolas, and parabolas. These curves appear throughout mathematics, physics, and engineering—from planetary orbits to satellite dishes.

Each conic section can be described by a second-degree polynomial equation in two variables:

\[Q(x, y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]

The coefficients \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) determine which type of conic section the equation represents.

1.2 Identifying Conic Sections

To determine what type of conic section an equation represents, we use the discriminant \(\Delta = B^2 - 4AC\):

  • If \(\Delta < 0\), the curve is an ellipse (or circle, which is a special case of an ellipse)
  • If \(\Delta = 0\), the curve is a parabola
  • If \(\Delta > 0\), the curve is a hyperbola

This classification is fundamental because it tells us the geometric nature of the curve before we even try to plot it.

1.3 Tangent and Normal Lines to Conic Sections
1.3.1 Finding Tangent Lines

A tangent line to a curve at a point touches the curve at exactly that point without crossing it (locally). The slope of the tangent line at a point \((x_0, y_0)\) equals the derivative \(\frac{dy}{dx}\) evaluated at that point.

For curves defined implicitly (where \(y\) is not isolated), we use implicit differentiation:

  1. Differentiate both sides of the equation with respect to \(x\)
  2. Apply the chain rule: when differentiating terms with \(y\), multiply by \(\frac{dy}{dx}\) (or \(y'\))
  3. Solve for \(\frac{dy}{dx}\)
  4. Substitute the point coordinates to find the specific slope

For example, for the ellipse \(3x^2 + 2y^2 = 5\):

\[\frac{d}{dx}(3x^2 + 2y^2 - 5) = 0\] \[6x + 4y \cdot y' = 0\] \[y' = -\frac{3x}{2y}\]

1.3.2 Finding Normal Lines

A normal line to a curve at a point is perpendicular to the tangent line at that point. If the tangent line has slope \(m_1\), the normal line has slope \(m_2\) where:

\[m_1 \cdot m_2 = -1 \quad \Rightarrow \quad m_2 = -\frac{1}{m_1}\]

Once we have the slope, we use the point-slope form to find the line equation:

\[y - y_0 = m(x - x_0)\]

1.4 Eccentricity of Conic Sections

Eccentricity (denoted \(\varepsilon\) or \(e\)) is a number that describes how “stretched” a conic section is. It provides a unified way to characterize all conic sections:

  • For a circle: \(\varepsilon = 0\)
  • For an ellipse: \(0 < \varepsilon < 1\)
  • For a parabola: \(\varepsilon = 1\)
  • For a hyperbola: \(\varepsilon > 1\)
1.4.1 Eccentricity of an Ellipse

For an ellipse with semi-major axis \(a\) and semi-minor axis \(b\) (where \(a > b\)), the eccentricity is:

\[\varepsilon = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}\]

where \(c\) is the distance from the center to each focus, related by \(c^2 = a^2 - b^2\).

As \(\varepsilon\) approaches 0, the ellipse becomes more circular. As \(\varepsilon\) approaches 1, it becomes more elongated.

1.4.2 Eccentricity of a Hyperbola

For a hyperbola with semi-transverse axis \(a\) and semi-conjugate axis \(b\), the eccentricity is:

\[\varepsilon = \frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}}\]

where \(c^2 = a^2 + b^2\) (note the plus sign, unlike ellipses).

1.5 Matrix Representation of Quadratic Forms

The general conic equation can be written in several equivalent forms using matrices:

Polynomial form: \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]

Matrix of quadratic form: \[\begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + \begin{bmatrix} D & E \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} + F = 0\]

Matrix of quadratic equation: \[\begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix} = 0\]

The matrix \(A_{qq} = \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix}\) is called the matrix of the quadratic form.

The matrix \(A_q = \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}\) is called the matrix of the quadratic equation.

1.6 Canonical Forms of Conic Sections

The canonical form of a conic section is its simplest equation, typically achieved by:

  1. Rotating the coordinate system to eliminate the \(xy\) term (making \(B = 0\))
  2. Translating the coordinate system to center the conic at the origin

For example:

  • Ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
  • Hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) or \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
  • Parabola: \(y^2 = 4px\) or \(x^2 = 4py\)
1.7 Rotation of Coordinate Systems

When a conic equation contains an \(xy\) term (i.e., \(B \neq 0\)), the conic is rotated relative to the coordinate axes. To eliminate this term, we rotate the coordinate system by an angle \(\theta\).

1.7.1 Finding the Rotation Angle

The rotation angle \(\theta\) is determined by:

\[\cot 2\theta = \frac{A - C}{B}\]

From this, we calculate:

\[\cos 2\theta = \frac{A - C}{\sqrt{(A - C)^2 + B^2}}\]

Then, using half-angle formulas:

  • If \(\cot 2\theta > 0\): \(\cos \theta = +\sqrt{\frac{1}{2}(1 + \cos 2\theta)}\)
  • If \(\cot 2\theta < 0\): \(\cos \theta = -\sqrt{\frac{1}{2}(1 + \cos 2\theta)}\)

Always: \(\sin \theta = +\sqrt{\frac{1}{2}(1 - \cos 2\theta)}\)

Special cases:

  • If \(B = 0\) and \(A < C\): \(\theta = 0\)
  • If \(B = 0\) and \(A > C\): \(\theta = \frac{\pi}{2}\)
  • If \(B \neq 0\) and \(A = C\): \(\theta = \frac{\pi}{4}\), so \(\sin \theta = \cos \theta = \frac{\sqrt{2}}{2}\)
1.7.2 Rotation Matrix

The transformation from old coordinates \((x, y)\) to new coordinates \((x', y')\) is:

\[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}\]

Or explicitly: \[x = x' \cos \theta - y' \sin \theta\] \[y = x' \sin \theta + y' \cos \theta\]

This rotation matrix is orthogonal, meaning its inverse equals its transpose.

1.8 Translation of Coordinate Systems

After rotation eliminates the \(xy\) term, there may still be linear terms (\(x'\) and \(y'\) terms). To eliminate these and center the conic at the origin, we translate the coordinate system.

If the rotated equation is: \[A'(x')^2 + C'(y')^2 + D'x' + E'y' + F' = 0\]

We complete the square to find the center \((h', k')\) and rewrite as: \[A'(x' - h')^2 + C'(y' - k')^2 = \text{constant}\]

Then substitute \(x'' = x' - h'\) and \(y'' = y' - k'\) to get the canonical form.

1.9 Method of Orthogonal Invariants

The method of orthogonal invariants provides a shortcut to find the canonical form without explicitly performing rotation and translation. This method uses quantities that remain unchanged (invariant) under rotation and translation.

1.9.1 Invariants for Ellipses and Hyperbolas

For non-parabolic conics, three key invariants are:

\[A + C = \tilde{A} + \tilde{C}\] \[\det \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix} = \det \begin{bmatrix} \tilde{A} & 0 \\ 0 & \tilde{C} \end{bmatrix} = \tilde{A} \cdot \tilde{C}\] \[\det \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix} = \det \begin{bmatrix} \tilde{A} & 0 & 0 \\ 0 & \tilde{C} & 0 \\ 0 & 0 & \tilde{F} \end{bmatrix} = \tilde{A} \cdot \tilde{C} \cdot \tilde{F}\]

where \(\tilde{A}\), \(\tilde{C}\), and \(\tilde{F}\) are the coefficients in the canonical form.

This gives us a system of three equations with three unknowns, which we can solve to find \(\tilde{A}\), \(\tilde{C}\), and \(\tilde{F}\).

1.9.2 Finding the Center

The center coordinates \((x_0, y_0)\) in the original coordinate system are found by solving:

\[\begin{cases} Ax_0 + \frac{B}{2}y_0 + \frac{D}{2} = 0 \\ \frac{B}{2}x_0 + Cy_0 + \frac{E}{2} = 0 \end{cases}\]

1.9.3 Invariants for Parabolas

For parabolas (\(B^2 - 4AC = 0\)), the method must be modified:

\[\tilde{C} = A + C\] \[\tilde{D} = 2\sqrt{\frac{-\Delta}{A + C}}\]

where \(\Delta = \det \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}\)

The canonical form is then: \[\tilde{C}y^2 + \tilde{D}x = 0\]

or equivalently: \[y^2 = 4 \cdot \frac{1}{2}\sqrt{\frac{\Delta}{(A+C)^3}} \cdot x\]

1.10 Parametric Equations

Parametric equations express the coordinates of points on a curve as functions of a parameter (usually \(\theta\) or \(t\)). This representation is particularly useful for circles and ellipses.

1.10.1 Circle Parametric Equations

For a circle centered at \((h, k)\) with radius \(r\):

\[x = h + r\cos\theta\] \[y = k + r\sin\theta\]

where \(\theta \in [0, 2\pi)\).

As \(\theta\) varies from \(0\) to \(2\pi\), the point \((x, y)\) traces the entire circle once.

1.10.2 Ellipse Parametric Equations

For an ellipse centered at \((h, k)\) with semi-major axis \(a\) (horizontal) and semi-minor axis \(b\) (vertical):

\[x = h + a\cos\theta\] \[y = k + b\sin\theta\]

where \(\theta \in [0, 2\pi)\).

This generalizes the circle equations by scaling differently in the \(x\) and \(y\) directions.

1.11 Implicit Differentiation and Partial Derivatives

For curves defined implicitly by \(F(x, y) = 0\), we can find \(\frac{dy}{dx}\) using partial derivatives.

1.11.1 Partial Derivatives

The partial derivative of \(F\) with respect to \(x\) (denoted \(F_x\) or \(\frac{\partial F}{\partial x}\)) is found by differentiating \(F\) with respect to \(x\) while treating \(y\) as a constant.

Similarly, \(F_y\) or \(\frac{\partial F}{\partial y}\) is found by differentiating with respect to \(y\) while treating \(x\) as constant.

1.11.2 General Implicit Differentiation Formula

For a curve \(F(x, y) = 0\), the derivative is:

\[\frac{dy}{dx} = -\frac{F_x}{F_y}\]

This formula is valid when \(F_y \neq 0\).

Why this works: By the chain rule, differentiating \(F(x, y) = 0\) with respect to \(x\) gives: \[F_x + F_y \cdot \frac{dy}{dx} = 0\]

Solving for \(\frac{dy}{dx}\) yields the formula above.

1.11.3 Applications

This general formula is extremely powerful:

  • Tangent lines: Once we have \(\frac{dy}{dx}\), we know the slope at any point
  • Normal lines: The normal slope is the negative reciprocal
  • Vertical tangents: Occur where \(F_y = 0\) (and \(F_x \neq 0\))
  • Horizontal tangents: Occur where \(F_x = 0\) (and \(F_y \neq 0\))
1.12 Key Properties of Specific Conics
1.12.1 Ellipse Properties

For an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (where \(a > b\)):

  • Center: \((0, 0)\)
  • Vertices: \((\pm a, 0)\)
  • Co-vertices: \((0, \pm b)\)
  • Foci: \((\pm c, 0)\) where \(c = \sqrt{a^2 - b^2}\)
  • Eccentricity: \(\varepsilon = \frac{c}{a}\)
1.12.2 Hyperbola Properties

For a hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\):

  • Center: \((0, 0)\)
  • Vertices: \((0, \pm a)\)
  • Foci: \((0, \pm c)\) where \(c = \sqrt{a^2 + b^2}\)
  • Asymptotes: \(y = \pm\frac{a}{b}x\)
  • Eccentricity: \(\varepsilon = \frac{c}{a}\)
1.12.3 Parabola Properties

For a parabola \(y^2 = 4px\):

  • Vertex: \((0, 0)\)
  • Focus: \((p, 0)\)
  • Directrix: \(x = -p\)
  • Axis of symmetry: \(y = 0\) (the \(x\)-axis)
  • Focal parameter: \(p\) (distance from vertex to focus)
1.13 Applied Problem-Solving Patterns
1.13.1 Completing Squares for Axis-Aligned Conics

To convert \(Ax^2 + Cy^2 + Dx + Ey + F = 0\) into standard form, group \(x\)-terms and \(y\)-terms, factor \(A\) and \(C\), then complete the square inside each group. Isolate the constant term on the opposite side before dividing to match the canonical ellipse or hyperbola templates. This workflow reveals the center \((h, k)\), denominators \(a^2\) and \(b^2\), and makes it effortless to compute eccentricity using \(e = \sqrt{1 - \frac{b^2}{a^2}}\) for ellipses or \(e = \sqrt{1 + \frac{b^2}{a^2}}\) for hyperbolas.

1.13.2 Focus-Directrix Construction for Parabolas

When given a focus \((x_f, y_f)\) and a directrix line, equate the squared distance from a general point \((x, y)\) to the focus with its squared distance to the directrix. Simplifying the resulting equation produces the familiar \((x - h)^2 = 4p(y - k)\) (vertical axis) or \((y - k)^2 = 4p(x - h)\) (horizontal axis) form, where \(p\) is the directed distance from the vertex to the focus.

1.13.3 Apollonius Circles for Distance Ratios

Loci defined by \(\frac{PA}{PB} = k\) (\(k > 0\), \(k \neq 1\)) are Apollonius circles. Setting \(\sqrt{(x - x_A)^2 + (y - y_A)^2} = k \sqrt{(x - x_B)^2 + (y - y_B)^2}\) and squaring both sides leads to a circle whose center lies on segment \(AB\) and whose radius depends on \(k\). This technique generalizes to any prescribed ratio.

1.13.4 Circumcircle Through Three Points

Any three non-collinear points determine a unique circumcircle. Substitute each point into the general circle \(x^2 + y^2 + Dx + Ey + F = 0\) to build a linear system for \(D\), \(E\), and \(F\). Solving yields the center \((-D/2, -E/2)\) and radius \(\sqrt{(D/2)^2 + (E/2)^2 - F}\).

1.13.5 Classifying Points with the Power of a Point

Rewrite the circle in center-radius form \((x - h)^2 + (y - k)^2 = r^2\). For any test point \((x_0, y_0)\), compute its power \(P = (x_0 - h)^2 + (y_0 - k)^2 - r^2\). The sign of \(P\) indicates whether the point lies inside (\(P < 0\)), on (\(P = 0\)), or outside (\(P > 0\)) the circle without graphing.

1.13.6 Recognizing Degenerate Quadratic Forms

If \(B^2 - 4AC = 0\) and the quadratic part factors as \((ux + vy)^2\), the conic is degenerate. After rotating the axes so that the \(xy\) term vanishes, the equation often reduces to \((x')^2 = c\) or \((x')(x'' ) = 0\), revealing coincident or parallel lines instead of a true curve.

2. Definitions

  • Conic Section: A curve obtained by intersecting a plane with a double cone; includes ellipses, hyperbolas, and parabolas.
  • Ellipse: A conic section where the sum of distances from any point on the curve to two fixed points (foci) is constant. Alternatively, a conic with eccentricity \(0 < \varepsilon < 1\).
  • Hyperbola: A conic section where the difference of distances from any point on the curve to two fixed points (foci) is constant. Alternatively, a conic with eccentricity \(\varepsilon > 1\).
  • Parabola: A conic section where any point on the curve is equidistant from a fixed point (focus) and a fixed line (directrix). Alternatively, a conic with eccentricity \(\varepsilon = 1\).
  • Discriminant: For the general conic equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant is \(\Delta = B^2 - 4AC\), which determines the type of conic.
  • Tangent Line: A line that touches a curve at exactly one point (locally) and has the same slope as the curve at that point.
  • Normal Line: A line perpendicular to the tangent line at a given point on a curve.
  • Eccentricity: A non-negative number that characterizes the shape of a conic section; denoted \(\varepsilon\) or \(e\).
  • Semi-major Axis: For an ellipse, half the length of the longest diameter; denoted \(a\).
  • Semi-minor Axis: For an ellipse, half the length of the shortest diameter; denoted \(b\).
  • Semi-transverse Axis: For a hyperbola, half the distance between the two vertices; denoted \(a\).
  • Semi-conjugate Axis: For a hyperbola, half the length perpendicular to the transverse axis used to construct the asymptotes; denoted \(b\).
  • Focus (plural: Foci): A special point (or points) associated with a conic section; for ellipses and hyperbolas, there are two foci.
  • Canonical Form: The simplest form of a conic equation, achieved by rotating and translating coordinates to eliminate the \(xy\) term and center the conic appropriately.
  • Quadratic Form: The homogeneous quadratic part of a conic equation, expressible as a matrix product: \(\begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\).
  • Matrix of Quadratic Form: The matrix \(A_{33} = \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix}\) that represents the quadratic part of a conic equation.
  • Matrix of Quadratic Equation: The matrix \(A_q = \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}\) that represents the entire conic equation.
  • Rotation Matrix: An orthogonal matrix \(\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\) used to rotate coordinate axes by angle \(\theta\).
  • Orthogonal Invariant: A quantity computed from the coefficients of a conic equation that remains unchanged under rotation and translation of coordinates.
  • Implicit Differentiation: A technique for finding \(\frac{dy}{dx}\) when \(y\) is not explicitly solved for, by differentiating both sides of an equation and applying the chain rule.
  • Partial Derivative: The derivative of a multivariable function with respect to one variable, treating all other variables as constants; denoted \(\frac{\partial F}{\partial x}\) or \(F_x\).
  • Parametric Equations: Equations that express coordinates as functions of a parameter (e.g., \(x = f(t)\), \(y = g(t)\)), allowing curves to be traced as the parameter varies.
  • Asymptote: A line that a curve approaches arbitrarily closely as it extends to infinity; hyperbolas have two asymptotes.
  • Directrix: A fixed line used in the definition of a parabola; every point on the parabola is equidistant from the focus and directrix.
  • Vertex: A special point on a conic section; for parabolas, it’s the point closest to the directrix; for ellipses and hyperbolas, they are the endpoints of the major/transverse axis.
  • Apollonius Circle: The locus of points whose distances to two fixed points are in a constant ratio; always a circle when the ratio is positive and not equal to \(1\).
  • Circumcircle: The unique circle passing through three non-collinear points, often constructed via perpendicular bisectors of two chords.
  • Power of a Point: The signed quantity \((x - h)^2 + (y - k)^2 - r^2\) relative to a circle \((x - h)^2 + (y - k)^2 = r^2\), indicating whether a point is inside, on, or outside the circle.
  • Degenerate Conic: A quadratic equation that factors into lines (intersecting, coincident, or parallel) instead of producing a non-degenerate curve such as an ellipse, parabola, or hyperbola.

3. Formulas

  • General Conic Equation: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)
  • Discriminant for Conic Type: \(\Delta = B^2 - 4AC\)
    • \(\Delta < 0\): ellipse
    • \(\Delta = 0\): parabola
    • \(\Delta > 0\): hyperbola
  • Slope of Tangent Line (Implicit Differentiation): For curve \(F(x, y) = 0\): \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{\partial F/\partial x}{\partial F/\partial y}\]
  • Perpendicularity Condition: Two lines with slopes \(m_1\) and \(m_2\) are perpendicular if and only if \(m_1 \cdot m_2 = -1\).
  • Point-Slope Form of Line: \(y - y_0 = m(x - x_0)\) where \(m\) is slope and \((x_0, y_0)\) is a point on the line.
  • Eccentricity of Ellipse: \(\varepsilon = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}\) where \(c^2 = a^2 - b^2\)
  • Eccentricity of Hyperbola: \(\varepsilon = \frac{c}{a} = \sqrt{1 + \frac{b^2}{a^2}}\) where \(c^2 = a^2 + b^2\)
  • Ellipse Canonical Form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (for \(a > b\))
  • Hyperbola Canonical Forms:
    • \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (horizontal transverse axis)
    • \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) (vertical transverse axis)
  • Parabola Canonical Forms:
    • \(y^2 = 4px\) (opens right/left)
    • \(x^2 = 4py\) (opens up/down)
  • Rotation Angle: \(\cot 2\theta = \frac{A - C}{B}\) and \(\cos 2\theta = \frac{A - C}{\sqrt{(A - C)^2 + B^2}}\)
  • Half-Angle Formulas:
    • \(\cos\theta = \pm\sqrt{\frac{1 + \cos 2\theta}{2}}\) (sign depends on \(\cot 2\theta\))
    • \(\sin\theta = +\sqrt{\frac{1 - \cos 2\theta}{2}}\) (always positive in standard method)
  • Rotation Matrix: \(\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}\)
  • Matrix of Quadratic Form: \(A_{33} = \begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix}\)
  • Matrix of Quadratic Equation: \(A_q = \begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix}\)
  • Orthogonal Invariants (Ellipse/Hyperbola):
    • \(A + C = \tilde{A} + \tilde{C}\)
    • \(\det(A_{33}) = \tilde{A} \cdot \tilde{C}\)
    • \(\det(A_q) = \tilde{A} \cdot \tilde{C} \cdot \tilde{F}\)
  • Center Coordinates: Solve the system \[\begin{cases} Ax_0 + \frac{B}{2}y_0 + \frac{D}{2} = 0 \\ \frac{B}{2}x_0 + Cy_0 + \frac{E}{2} = 0 \end{cases}\]
  • Orthogonal Invariants (Parabola):
    • \(\tilde{C} = A + C\)
    • \(\tilde{D} = 2\sqrt{\frac{-\Delta}{A + C}}\) where \(\Delta = \det(A_q)\)
  • Circle Parametric Equations: \(x = h + r\cos\theta\), \(y = k + r\sin\theta\) for center \((h, k)\) and radius \(r\)
  • Ellipse Parametric Equations: \(x = h + a\cos\theta\), \(y = k + b\sin\theta\) for center \((h, k)\), semi-axes \(a\) and \(b\)
  • Distance Formula: \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
  • Tangent Length from External Point: For a circle with center \(C\), radius \(r\), and external point \(A\): \[\text{Tangent length} = \sqrt{(AC)^2 - r^2}\]
  • Hyperbola Asymptotes (for \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\)): \(y = \pm\frac{a}{b}x\)
  • Pythagorean Identity: \(\cos^2\theta + \sin^2\theta = 1\)
  • Apollonius Circle Equation: For points \(A\) and \(B\) with ratio \(k\), the locus is \((x - x_A)^2 + (y - y_A)^2 = k^2\big[(x - x_B)^2 + (y - y_B)^2\big]\), which simplifies to a circle when \(k > 0\) and \(k \neq 1\).
  • Circle Through Three Points (Determinant Form): \[\begin{vmatrix} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0\]
  • Point Classification (Power Test): For circle \((x - h)^2 + (y - k)^2 = r^2\), compute \(P = (x_0 - h)^2 + (y_0 - k)^2 - r^2\). Then \(P < 0\) (inside), \(P = 0\) (on), \(P > 0\) (outside).
  • Focus-Directrix Parabola: Equate \((x - x_f)^2 + (y - y_f)^2 = \frac{(ax + by + c)^2}{a^2 + b^2}\) for directrix \(ax + by + c = 0\) to obtain a quadratic that simplifies to \((x - h)^2 = 4p(y - k)\) or \((y - k)^2 = 4p(x - h)\) after completing the square.
  • Degenerate Pair of Lines: If \(Ax^2 + Bxy + Cy^2 + F = 0\) has \(B^2 - 4AC = 0\) and no linear terms, rotating the axes by \(\theta\) with \(\cot 2\theta = \frac{A - C}{B}\) yields \(\lambda (x')^2 = -F\), which factors as \(x' = \pm\sqrt{-F/\lambda}\), i.e., two parallel lines.

4. Examples

4.1. Partial Derivatives Practice (Lab 8, Task 1)

Find \(F_x\) and \(F_y\) for \(F(x, y) = x^3 + 3x^2y - y^3\).

Click to see the solution

Key Concept: Partial derivatives treat one variable at a time, holding others constant.

  1. Find \(F_x\) (partial derivative with respect to \(x\)):

    Treat \(y\) as a constant: \[F_x = \frac{\partial F}{\partial x} = \frac{\partial}{\partial x}(x^3 + 3x^2y - y^3)\] \[F_x = 3x^2 + 6xy - 0\] \[F_x = 3x^2 + 6xy\]

  2. Find \(F_y\) (partial derivative with respect to \(y\)):

    Treat \(x\) as a constant: \[F_y = \frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(x^3 + 3x^2y - y^3)\] \[F_y = 0 + 3x^2 - 3y^2\] \[F_y = 3x^2 - 3y^2\]

Answer:

\(F_x = 3x^2 + 6xy\)

\(F_y = 3x^2 - 3y^2\)

4.2. Partial Derivatives with Trigonometric Functions (Lab 8, Task 2)

Find \(F_x\) and \(F_y\) for \(F(x, y) = \sin(x) + \cos(y)\).

Click to see the solution

  1. Find \(F_x\):

    \[F_x = \frac{\partial}{\partial x}(\sin(x) + \cos(y)) = \cos(x) + 0 = \cos(x)\]

  2. Find \(F_y\):

    \[F_y = \frac{\partial}{\partial y}(\sin(x) + \cos(y)) = 0 - \sin(y) = -\sin(y)\]

Answer:

\(F_x = \cos(x)\)

\(F_y = -\sin(y)\)

4.3. General Formula for Derivative (Lab 8, Task 3)

Find \(\frac{dy}{dx}\) for the circle \(x^2 + y^2 = 25\) using the general formula \(\frac{dy}{dx} = -\frac{F_x}{F_y}\).

Click to see the solution

Key Concept: Apply the general implicit differentiation formula with partial derivatives.

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = x^2 + y^2 - 25 = 0\]

  2. Find partial derivatives:

    \[F_x = \frac{\partial F}{\partial x} = 2x\] \[F_y = \frac{\partial F}{\partial y} = 2y\]

  3. Apply the formula:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{2x}{2y} = -\frac{x}{y}\]

Answer: \(\frac{dy}{dx} = -\frac{x}{y}\)

4.4. General Formula Applied to Ellipse (Lab 8, Task 4)

Find \(\frac{dy}{dx}\) for the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) using the general formula.

Click to see the solution

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = \frac{x^2}{9} + \frac{y^2}{4} - 1 = 0\]

  2. Find partial derivatives:

    \[F_x = \frac{\partial F}{\partial x} = \frac{2x}{9}\] \[F_y = \frac{\partial F}{\partial y} = \frac{2y}{4} = \frac{y}{2}\]

  3. Apply the formula:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{2x/9}{y/2} = -\frac{2x}{9} \cdot \frac{2}{y} = -\frac{4x}{9y}\]

Answer: \(\frac{dy}{dx} = -\frac{4x}{9y}\)

4.5. Tangent to Complex Curve (Lab 8, Task 5)

Find the slope of the tangent line to the curve \(x^3 + y^3 = 6xy\) at the point \((3, 3)\).

Click to see the solution

Key Concept: Use the general implicit differentiation formula for complex equations.

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = x^3 + y^3 - 6xy = 0\]

  2. Find partial derivatives:

    \[F_x = \frac{\partial F}{\partial x} = 3x^2 - 6y\] \[F_y = \frac{\partial F}{\partial y} = 3y^2 - 6x\]

  3. Apply the formula:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{3x^2 - 6y}{3y^2 - 6x}\]

  4. Evaluate at point \((3, 3)\):

    \[\frac{dy}{dx}\bigg|_{(3,3)} = -\frac{3(3)^2 - 6(3)}{3(3)^2 - 6(3)} = -\frac{27 - 18}{27 - 18} = -\frac{9}{9} = -1\]

  5. Tangent line equation:

    Using point-slope form: \[y - 3 = -1(x - 3)\] \[y = -x + 6\]

Answer: Slope: \(m = -1\); Tangent line: \(y = -x + 6\)

4.6. General Formula Practice (Lab 8, Task 6)

Use the general formula to find \(\frac{dy}{dx}\) for the curve \(\sin(x) + \cos(y) = 1\).

Click to see the solution

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = \sin(x) + \cos(y) - 1 = 0\]

  2. Find partial derivatives:

    \[F_x = \cos(x)\] \[F_y = -\sin(y)\]

  3. Apply the formula:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{\cos(x)}{-\sin(y)} = \frac{\cos(x)}{\sin(y)}\]

Answer: \(\frac{dy}{dx} = \frac{\cos(x)}{\sin(y)}\)

4.7. Find Slope at a Point (Lab 8, Task 7)

Find the slope of the tangent line to \(x^2y + y^2x = 2\) at the point \((1, 1)\).

Click to see the solution

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = x^2y + y^2x - 2 = 0\]

  2. Find partial derivatives:

    \[F_x = \frac{\partial}{\partial x}(x^2y + y^2x - 2) = 2xy + y^2\] \[F_y = \frac{\partial}{\partial y}(x^2y + y^2x - 2) = x^2 + 2yx\]

  3. Apply the formula:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{2xy + y^2}{x^2 + 2yx}\]

  4. Evaluate at \((1, 1)\):

    \[\frac{dy}{dx}\bigg|_{(1,1)} = -\frac{2(1)(1) + (1)^2}{(1)^2 + 2(1)(1)} = -\frac{2 + 1}{1 + 2} = -\frac{3}{3} = -1\]

Answer: \(m = -1\)

4.8. Vertical Tangents (Lab 8, Task 8)

Find the points on the curve \(x^2 + y^3 - 3xy = 0\) where the tangent is vertical.

Click to see the solution

Key Concept: Vertical tangents occur where \(F_y = 0\) (denominator of \(\frac{dy}{dx}\) is zero).

  1. Write as \(F(x, y) = 0\):

    \[F(x, y) = x^2 + y^3 - 3xy = 0\]

  2. Find partial derivatives:

    \[F_x = 2x - 3y\] \[F_y = 3y^2 - 3x\]

  3. Vertical tangent occurs when \(F_y = 0\):

    \[3y^2 - 3x = 0\] \[x = y^2\]

  4. Substitute into original equation:

    \[(y^2)^2 + y^3 - 3(y^2)y = 0\] \[y^4 + y^3 - 3y^3 = 0\] \[y^4 - 2y^3 = 0\] \[y^3(y - 2) = 0\]

    So \(y = 0\) or \(y = 2\).

  5. Find corresponding \(x\) values:

    • When \(y = 0\): \(x = (0)^2 = 0\) → Point \((0, 0)\)
    • When \(y = 2\): \(x = (2)^2 = 4\) → Point \((4, 2)\)

Answer: Vertical tangents at \((0, 0)\) and \((4, 2)\)

4.9. Parametric Curve Identification (Lecture 8, Example 1)

As \(\theta\) varies from \(0\) to \(2\pi\), the point \(M(2 + 3\cos\theta, 1 + 3\sin\theta)\) traces which curve?

Click to see the solution

Key Concept: Recognize parametric equations and use the Pythagorean identity to eliminate the parameter.

  1. Write the parametric equations:

    \[x = 2 + 3\cos\theta\] \[y = 1 + 3\sin\theta\]

    where \(\theta \in [0, 2\pi)\).

  2. Isolate trigonometric functions:

    From the parametric equations: \[\cos\theta = \frac{x - 2}{3}\] \[\sin\theta = \frac{y - 1}{3}\]

  3. Use Pythagorean identity:

    We know that \(\cos^2\theta + \sin^2\theta = 1\). Substituting: \[\left(\frac{x - 2}{3}\right)^2 + \left(\frac{y - 1}{3}\right)^2 = 1\]

    \[\frac{(x - 2)^2}{9} + \frac{(y - 1)^2}{9} = 1\]

    Multiplying both sides by 9: \[(x - 2)^2 + (y - 1)^2 = 9\]

  4. Identify the curve:

    This is the equation of a circle in standard form \((x - h)^2 + (y - k)^2 = r^2\) where:

    • Center: \((h, k) = (2, 1)\)
    • Radius: \(r = \sqrt{9} = 3\)

Answer: The point traces a circle with center \((2, 1)\) and radius \(3\). Equation: \((x - 2)^2 + (y - 1)^2 = 9\)

4.10. Tangent Length to Circle (Lecture 8, Example 2)

Find the length of the tangent drawn from the point \(A(4, 3)\) to the circle given by the equation \(x^2 + y^2 - 2x - 4y + 1 = 0\).

Click to see the solution

Key Concept: Convert the circle to standard form, then use the Pythagorean theorem with the distance from the external point to the center.

  1. Identify the circle’s center and radius:

    Given: \(x^2 + y^2 - 2x - 4y + 1 = 0\)

    Complete the square: \[(x^2 - 2x) + (y^2 - 4y) + 1 = 0\] \[(x^2 - 2x + 1) + (y^2 - 4y + 4) + 1 - 1 - 4 = 0\] \[(x - 1)^2 + (y - 2)^2 - 4 = 0\] \[(x - 1)^2 + (y - 2)^2 = 4\]

    So:

    • Center: \(C = (1, 2)\)
    • Radius: \(r = \sqrt{4} = 2\)

  2. Calculate distance from point to center:

    Point \(A = (4, 3)\), Center \(C = (1, 2)\): \[AC = \sqrt{(4 - 1)^2 + (3 - 2)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}\]

  3. Apply tangent length formula:

    The tangent from an external point to a circle forms a right angle with the radius at the point of tangency. Using the Pythagorean theorem: \[\text{Tangent length}^2 + r^2 = AC^2\] \[\text{Tangent length} = \sqrt{AC^2 - r^2}\]

    Substituting: \[\text{Tangent length} = \sqrt{(\sqrt{10})^2 - 2^2} = \sqrt{10 - 4} = \sqrt{6}\]

Answer: \(\sqrt{6}\)

4.11. Tangent Line to Circle (Lecture 8, Example 3)

Find the equation of the tangent line to the circle given by \(x^2 + y^2 + 4x - 2y - 20 = 0\) at the point \(P(2, 4)\).

Click to see the solution

Key Concept: Use implicit differentiation to find the slope at the given point, then apply point-slope form.

  1. Verify the point lies on the circle:

    Substitute \(x = 2\), \(y = 4\): \[(2)^2 + (4)^2 + 4(2) - 2(4) - 20 = 4 + 16 + 8 - 8 - 20 = 0\] ✓

  2. Implicit differentiation:

    Differentiate both sides with respect to \(x\): \[\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(2y) - \frac{d}{dx}(20) = 0\] \[2x + 2y \cdot \frac{dy}{dx} + 4 - 2 \cdot \frac{dy}{dx} = 0\]

  3. Solve for \(\frac{dy}{dx}\):

    \[2y \cdot \frac{dy}{dx} - 2 \cdot \frac{dy}{dx} = -2x - 4\] \[2(y - 1) \cdot \frac{dy}{dx} = -2(x + 2)\] \[\frac{dy}{dx} = -\frac{x + 2}{y - 1}\]

  4. Find the slope at \(P(2, 4)\):

    \[m = \frac{dy}{dx}\bigg|_{(2,4)} = -\frac{2 + 2}{4 - 1} = -\frac{4}{3}\]

  5. Equation of the tangent line:

    Using point-slope form: \[y - 4 = -\frac{4}{3}(x - 2)\] \[y - 4 = -\frac{4}{3}x + \frac{8}{3}\] \[y = -\frac{4}{3}x + \frac{8}{3} + 4\] \[y = -\frac{4}{3}x + \frac{8}{3} + \frac{12}{3}\] \[y = -\frac{4}{3}x + \frac{20}{3}\]

Answer: \(y = -\frac{4}{3}x + \frac{20}{3}\)

4.12. Normal Line to Ellipse (Lecture 8, Example 4)

The point \(P(3, 2)\) lies on the ellipse defined by \(4x^2 + 9y^2 = 72\).

  1. Show that \(P\) lies on the ellipse.
  2. Find the slope of the normal line to the ellipse at point \(P\).
  3. Hence, find the equation of the normal line.
Click to see the solution

Key Concept: Use implicit differentiation to find the tangent slope, then use perpendicularity to find the normal slope.

(a) Verify the point:

Substitute \(x = 3\), \(y = 2\) into the equation: \[4(3)^2 + 9(2)^2 = 4(9) + 9(4) = 36 + 36 = 72\] ✓

The point \(P(3, 2)\) lies on the ellipse.

(b) Find slope of the normal:

  1. Implicit differentiation:

    \[\frac{d}{dx}(4x^2) + \frac{d}{dx}(9y^2) = \frac{d}{dx}(72)\] \[8x + 18y \cdot \frac{dy}{dx} = 0\] \[18y \cdot \frac{dy}{dx} = -8x\] \[\frac{dy}{dx} = -\frac{4x}{9y}\]

  2. Slope of tangent at \(P(3, 2)\):

    \[m_{\text{tangent}} = -\frac{4(3)}{9(2)} = -\frac{12}{18} = -\frac{2}{3}\]

  3. Slope of normal:

    \[m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-2/3} = \frac{3}{2}\]

(c) Equation of the normal line:

Using point-slope form with \(m = \frac{3}{2}\) and point \(P(3, 2)\): \[y - 2 = \frac{3}{2}(x - 3)\] \[y - 2 = \frac{3}{2}x - \frac{9}{2}\] \[y = \frac{3}{2}x - \frac{9}{2} + 2\] \[y = \frac{3}{2}x - \frac{9}{2} + \frac{4}{2}\] \[y = \frac{3}{2}x - \frac{5}{2}\]

Answer:

  1. Verified: \(P(3, 2)\) lies on the ellipse
  2. Slope of normal: \(m = \frac{3}{2}\)
  3. Normal line: \(y = \frac{3}{2}x - \frac{5}{2}\)
4.13. Tangent to Hyperbola (Lecture 8, Example 5)

Consider the rectangular hyperbola given by the equation \(xy = 12\).

  1. Use implicit differentiation to find an expression for \(\frac{dy}{dx}\).
  2. Find the equation of the tangent line to the hyperbola at the point \(P(3, 4)\).
  3. Show that this tangent line intersects the \(x\)-axis at \((6, 0)\).
Click to see the solution

Key Concept: Apply the product rule during implicit differentiation.

(a) Implicit differentiation:

\[\frac{d}{dx}(xy) = \frac{d}{dx}(12)\]

Using the product rule on the left side: \[x \cdot \frac{dy}{dx} + y \cdot 1 = 0\] \[x \cdot \frac{dy}{dx} = -y\] \[\frac{dy}{dx} = -\frac{y}{x}\]

(b) Equation of tangent line:

  1. Verify point \(P(3, 4)\) lies on hyperbola:

    \[3 \times 4 = 12\] ✓

  2. Find slope at \(P(3, 4)\):

    \[m = \frac{dy}{dx}\bigg|_{(3,4)} = -\frac{4}{3}\]

  3. Equation of tangent line:

    Using point-slope form: \[y - 4 = -\frac{4}{3}(x - 3)\] \[y - 4 = -\frac{4}{3}x + 4\] \[y = -\frac{4}{3}x + 8\]

(c) Show intersection with \(x\)-axis at \((6, 0)\):

Set \(y = 0\) in the tangent line equation: \[0 = -\frac{4}{3}x + 8\] \[\frac{4}{3}x = 8\] \[x = 8 \cdot \frac{3}{4} = 6\]

So the tangent line intersects the \(x\)-axis at \((6, 0)\). ✓

Answer:

  1. \(\frac{dy}{dx} = -\frac{y}{x}\)
  2. Tangent line: \(y = -\frac{4}{3}x + 8\)
  3. Verified: Intersection at \((6, 0)\)
4.14. Tangent and Normal Lines to an Ellipse (Lecture 8, Example 6)

Find the equations of the tangent and normal lines to the ellipse \(3x^2 + 2y^2 = 5\) at point \(Q(-1, 1)\).

Click to see the solution

Key Concept: Use implicit differentiation to find the slope of the tangent line, then use the perpendicularity condition to find the normal line slope.

  1. Find the derivative using implicit differentiation:

    Starting with \(3x^2 + 2y^2 = 5\), rewrite as: \[3x^2 + 2y^2 - 5 = 0\]

    Differentiate both sides with respect to \(x\): \[\frac{d}{dx}(3x^2) + \frac{d}{dx}(2y^2) - \frac{d}{dx}(5) = 0\] \[6x + 4y \cdot y' - 0 = 0\] \[6x + 4yy' = 0\]

    Solve for \(y'\): \[y' = -\frac{6x}{4y} = -\frac{3x}{2y}\]

  2. Find the slope of the tangent line at \(Q(-1, 1)\):

    \[m_{\text{tangent}} = y'(-1, 1) = -\frac{3(-1)}{2(1)} = \frac{3}{2}\]

  3. Find the equation of the tangent line:

    Using point-slope form with \(m = \frac{3}{2}\) and point \((-1, 1)\): \[y - 1 = \frac{3}{2}(x - (-1))\] \[y - 1 = \frac{3}{2}(x + 1)\] \[y - 1 = \frac{3}{2}x + \frac{3}{2}\] \[y = \frac{3}{2}x + \frac{5}{2}\]

  4. Find the slope of the normal line:

    The normal line is perpendicular to the tangent, so: \[m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{3/2} = -\frac{2}{3}\]

  5. Find the equation of the normal line:

    Using point-slope form with \(m = -\frac{2}{3}\) and point \((-1, 1)\): \[y - 1 = -\frac{2}{3}(x + 1)\] \[y - 1 = -\frac{2}{3}x - \frac{2}{3}\] \[y = -\frac{2}{3}x + \frac{1}{3}\]

Answer:

Tangent line: \(y = \frac{3}{2}x + \frac{5}{2}\)

Normal line: \(y = -\frac{2}{3}x + \frac{1}{3}\)

4.15. Eccentricity from Geometric Constraint (Lecture 8, Example 7)

Find the eccentricity of an ellipse given that its major axis subtends an angle of \(120°\) at the endpoints of its minor axis.

Click to see the solution

Key Concept: Use geometry to relate the major and minor axes, then apply the eccentricity formula.

  1. Set up the geometric relationship:

    Consider one quarter of the ellipse. The endpoints of the major axis and one endpoint of the minor axis form a triangle. The angle at the minor axis endpoint is half of \(120°\), which is \(60°\).

    Let \(a\) be the semi-major axis and \(b\) be the semi-minor axis. The full major axis has length \(2a\), and we’re looking at the triangle formed by:

    • One endpoint of the minor axis (at distance \(b\) from center)
    • Two endpoints of the major axis (at distance \(a\) from center each)

    The angle of \(120°\) is split into two \(60°\) angles.

  2. Apply trigonometry:

    In the right triangle formed, we have:

    • One leg of length \(a\) (semi-major axis)
    • Hypotenuse from minor axis endpoint to major axis endpoint

    Using the \(60°\) angle, we can show that: \[\tan 60° = \frac{a}{b}\] \[\sqrt{3} = \frac{a}{b}\] \[a = \sqrt{3} \cdot b\]

  3. Calculate eccentricity:

    For an ellipse, \(c^2 = a^2 - b^2\) where \(c\) is the focal distance.

    The eccentricity is: \[\varepsilon = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}}\]

    Substituting \(a = \sqrt{3}b\): \[\varepsilon = \sqrt{1 - \frac{b^2}{(\sqrt{3}b)^2}} = \sqrt{1 - \frac{b^2}{3b^2}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}\]

Answer: \(\varepsilon = \sqrt{\frac{2}{3}} \approx 0.816\)

4.16. Hyperbola Transformation to Canonical Form (Lecture 8, Example 8)

Prove that the curve given by \(7x^2 + 48xy - 7y^2 - 62x - 34y + 98 = 0\) is a hyperbola. Find the eccentricity of this hyperbola, coordinates of its center and foci. Find the equations of axes, asymptotes, and directrices of this hyperbola.

Click to see the solution

Key Concept: Use the discriminant to identify the conic type, then apply rotation and translation transformations to find the canonical form and properties.

  1. Identify coefficients:

    From \(7x^2 + 48xy - 7y^2 - 62x - 34y + 98 = 0\): \[A = 7, \quad B = 48, \quad C = -7, \quad D = -62, \quad E = -34, \quad F = 98\]

  2. Verify it’s a hyperbola:

    \[\Delta = B^2 - 4AC = 48^2 - 4(7)(-7) = 2304 + 196 = 2500 > 0\]

    Since \(\Delta > 0\), this is indeed a hyperbola.

  3. Calculate rotation angle:

    \[\cot 2\theta = \frac{A - C}{B} = \frac{7 - (-7)}{48} = \frac{14}{48} = \frac{7}{24}\]

    Since \(\cot 2\theta > 0\): \[\cos 2\theta = \frac{A - C}{\sqrt{(A - C)^2 + B^2}} = \frac{14}{\sqrt{14^2 + 48^2}} = \frac{14}{\sqrt{196 + 2304}} = \frac{14}{\sqrt{2500}} = \frac{14}{50} = \frac{7}{25}\]

    Using half-angle formulas (with positive sign since \(\cot 2\theta > 0\)): \[\cos\theta = +\sqrt{\frac{1 + \cos 2\theta}{2}} = \sqrt{\frac{1 + 7/25}{2}} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{32}{50}} = \sqrt{\frac{16}{25}} = \frac{4}{5}\]

    \[\sin\theta = +\sqrt{\frac{1 - \cos 2\theta}{2}} = \sqrt{\frac{1 - 7/25}{2}} = \sqrt{\frac{18/25}{2}} = \sqrt{\frac{18}{50}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\]

  4. Set up rotation transformation:

    \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ 3/5 & 4/5 \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}\]

    This gives: \[x = \frac{1}{5}(4x' - 3y'), \quad y = \frac{1}{5}(3x' + 4y')\]

  5. Substitute and simplify:

    After substituting into the original equation and performing matrix multiplication (or expanding algebraically), we get: \[25(x')^2 - 25(y')^2 - 70x' + 10y' + 98 = 0\]

  6. Complete the square:

    \[25(x')^2 - 70x' - 25(y')^2 + 10y' + 98 = 0\] \[25(x' - \frac{7}{5})^2 - 25 \cdot \frac{49}{25} - 25(y' - \frac{1}{5})^2 + 25 \cdot \frac{1}{25} + 98 = 0\] \[25(x' - \frac{7}{5})^2 - 25(y' - \frac{1}{5})^2 = 49 - 1 - 98 = -50\]

    Dividing by \(-50\): \[\frac{(y' - 1/5)^2}{2} - \frac{(x' - 7/5)^2}{2} = 1\]

  7. Apply translation:

    Let \(x'' = x' - \frac{7}{5}\) and \(y'' = y' - \frac{1}{5}\). The canonical form is: \[\frac{(y'')^2}{(\sqrt{2})^2} - \frac{(x'')^2}{(\sqrt{2})^2} = 1\]

    This is a hyperbola with \(a = b = \sqrt{2}\).

  8. Find properties in canonical form:

    • Eccentricity: \(\varepsilon = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{2}} = \sqrt{2}\)
    • Center in \(x''y''\) frame: \((0, 0)\)
    • Foci in \(x''y''\) frame: \((0, \pm c)\) where \(c = \sqrt{a^2 + b^2} = \sqrt{2 + 2} = 2\), so \((0, \pm 2)\)

  9. Transform back to original coordinates:

    The center in the \(x'y'\) frame is \((\frac{7}{5}, \frac{1}{5})\). Transform to \(xy\) frame: \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ 3/5 & 4/5 \end{bmatrix} \begin{bmatrix} 7/5 \\ 1/5 \end{bmatrix} = \begin{bmatrix} (28 - 3)/25 \\ (21 + 4)/25 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\]

    For the foci, add \((0, \pm 2)\) in the \(x''y''\) frame to the center in \(x'y'\) frame, then transform: \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 4/5 & -3/5 \\ 3/5 & 4/5 \end{bmatrix} \begin{bmatrix} 7/5 \\ 1/5 \pm 2 \end{bmatrix}\]

    This gives foci at approximately: \[F_1 = (1, 1) + (-\frac{6}{5}, \frac{8}{5}) = (-\frac{1}{5}, \frac{13}{5})\] \[F_2 = (1, 1) - (-\frac{6}{5}, \frac{8}{5}) = (\frac{11}{5}, -\frac{3}{5})\]

Answer:

  • Type: Hyperbola (confirmed)
  • Eccentricity: \(\varepsilon = \sqrt{2}\)
  • Center: \((1, 1)\)
  • Foci: \((-\frac{1}{5}, \frac{13}{5})\) and \((\frac{11}{5}, -\frac{3}{5})\)
  • Canonical form: \(\frac{y^2}{2} - \frac{x^2}{2} = 1\) (in the rotated and translated frame)
4.17. Hyperbola Using Orthogonal Invariants (Lecture 8, Example 8 Alternative)

Use the method of orthogonal invariants to find the canonical form of \(7x^2 + 48xy - 7y^2 - 62x - 34y + 98 = 0\).

Click to see the solution

Key Concept: Orthogonal invariants provide a shortcut to the canonical form without explicitly performing rotation.

  1. Identify coefficients:

    \[A = 7, \quad B = 48, \quad C = -7, \quad D = -62, \quad E = -34, \quad F = 98\]

  2. Verify it’s a hyperbola:

    \[B^2 - 4AC = 48^2 + 4(7)(7) = 2304 + 196 = 2500 > 0\]

  3. Set up the system of orthogonal invariants:

    \[\begin{cases} \tilde{A} + \tilde{C} = A + C = 7 + (-7) = 0 \\ \tilde{A} \cdot \tilde{C} = \det\begin{bmatrix} A & B/2 \\ B/2 & C \end{bmatrix} = \det\begin{bmatrix} 7 & 24 \\ 24 & -7 \end{bmatrix} = -49 - 576 = -625 \\ \tilde{A} \cdot \tilde{C} \cdot \tilde{F} = \det\begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix} \end{cases}\]

    Calculate the 3×3 determinant: \[\det\begin{bmatrix} 7 & 24 & -31 \\ 24 & -7 & -17 \\ -31 & -17 & 98 \end{bmatrix} = 7(-7 \cdot 98 - (-17)(-17)) + \ldots = -31250\]

    So \(\tilde{A} \cdot \tilde{C} \cdot \tilde{F} = -31250\).

  4. Solve the system:

    From \(\tilde{A} + \tilde{C} = 0\), we have \(\tilde{C} = -\tilde{A}\).

    Substituting into \(\tilde{A} \cdot \tilde{C} = -625\): \[\tilde{A} \cdot (-\tilde{A}) = -625\] \[-(\tilde{A})^2 = -625\] \[(\tilde{A})^2 = 625\] \[\tilde{A} = \pm 25\]

    Taking \(\tilde{A} = 25\) and \(\tilde{C} = -25\) (the choice doesn’t affect the final canonical form).

    From \(\tilde{A} \cdot \tilde{C} \cdot \tilde{F} = -31250\): \[25 \cdot (-25) \cdot \tilde{F} = -31250\] \[-625\tilde{F} = -31250\] \[\tilde{F} = 50\]

  5. Write the transformed equation:

    \[25x^2 - 25y^2 + 50 = 0\] \[25x^2 - 25y^2 = -50\] \[\frac{y^2}{2} - \frac{x^2}{2} = 1\]

  6. Find the center:

    Solve: \[\begin{cases} Ax_0 + \frac{B}{2}y_0 + \frac{D}{2} = 0 \\ \frac{B}{2}x_0 + Cy_0 + \frac{E}{2} = 0 \end{cases}\]

    \[\begin{cases} 7x_0 + 24y_0 - 31 = 0 \\ 24x_0 - 7y_0 - 17 = 0 \end{cases}\]

    Solving this system yields \(x_0 = 1\), \(y_0 = 1\).

Answer:

  • Canonical form: \(\frac{y^2}{2} - \frac{x^2}{2} = 1\)
  • Center in original coordinates: \((1, 1)\)
4.18. Parabola Transformation to Canonical Form (Lecture 8, Example 9)

Prove that the curve given by \(x^2 + 2xy + y^2 + x = 0\) is a parabola. Find the coordinates of its vertex and focus. Find the equations of its axis and directrix.

Click to see the solution

Key Concept: For parabolas, the rotation angle is special (\(\theta = 45°\) when \(A = C\)), and we need to complete the square carefully.

  1. Identify coefficients:

    \[A = 1, \quad B = 2, \quad C = 1, \quad D = 1, \quad E = 0, \quad F = 0\]

  2. Verify it’s a parabola:

    \[\Delta = B^2 - 4AC = 4 - 4(1)(1) = 0\]

    Since \(\Delta = 0\), this is indeed a parabola.

  3. Find rotation angle:

    \[\cot 2\theta = \frac{A - C}{B} = \frac{1 - 1}{2} = 0\]

    This means \(2\theta = 90°\), so \(\theta = 45°\).

    Therefore: \[\cos\theta = \sin\theta = \frac{\sqrt{2}}{2}\]

  4. Set up rotation transformation:

    \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{bmatrix} \begin{bmatrix} x' \\ y' \end{bmatrix}\]

    \[x = \frac{\sqrt{2}}{2}(x' - y'), \quad y = \frac{\sqrt{2}}{2}(x' + y')\]

  5. Substitute into original equation:

    \[\left(\frac{\sqrt{2}}{2}(x' - y')\right)^2 + 2\left(\frac{\sqrt{2}}{2}(x' - y')\right)\left(\frac{\sqrt{2}}{2}(x' + y')\right) + \left(\frac{\sqrt{2}}{2}(x' + y')\right)^2 + \frac{\sqrt{2}}{2}(x' - y') = 0\]

    Expanding and simplifying: \[\frac{1}{2}(x' - y')^2 + \frac{1}{2}(x'^2 - y'^2) + \frac{1}{2}(x' + y')^2 + \frac{\sqrt{2}}{2}(x' - y') = 0\]

    \[\frac{1}{2}(x'^2 - 2x'y' + y'^2) + \frac{1}{2}(x'^2 - y'^2) + \frac{1}{2}(x'^2 + 2x'y' + y'^2) + \frac{\sqrt{2}}{2}x' - \frac{\sqrt{2}}{2}y' = 0\]

    \[(x')^2 + \frac{\sqrt{2}}{2}x' - \frac{\sqrt{2}}{2}y' = 0\]

    Completing the square: \[(x' + \frac{\sqrt{2}}{8})^2 = \frac{\sqrt{2}}{4}(y' + \frac{\sqrt{2}}{16})\]

  6. Apply translation:

    Let \(x'' = x' + \frac{\sqrt{2}}{8}\) and \(y'' = y' + \frac{\sqrt{2}}{16}\).

    The canonical form becomes: \[(x'')^2 = \frac{\sqrt{2}}{4}y''\]

    or \[(x'')^2 = 4 \cdot \frac{\sqrt{2}}{16} \cdot y''\]

    So \(p = \frac{\sqrt{2}}{16}\) (the focal parameter).

  7. Find vertex and focus in canonical form:

    • Vertex: \((0, 0)\) in the \(x''y''\) frame
    • Focus: \((0, \frac{\sqrt{2}}{16})\) in the \(x''y''\) frame

  8. Transform back to original coordinates:

    The vertex in the \(x'y'\) frame is \((-\frac{\sqrt{2}}{8}, -\frac{\sqrt{2}}{16})\).

    Transform to \(xy\) frame: \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{bmatrix} \begin{bmatrix} -\sqrt{2}/8 \\ -\sqrt{2}/16 \end{bmatrix}\]

    \[= \begin{bmatrix} \frac{\sqrt{2}}{2} \cdot \frac{-\sqrt{2}}{8} + \frac{-\sqrt{2}}{2} \cdot \frac{-\sqrt{2}}{16} \\ \frac{\sqrt{2}}{2} \cdot \frac{-\sqrt{2}}{8} + \frac{\sqrt{2}}{2} \cdot \frac{-\sqrt{2}}{16} \end{bmatrix} = \begin{bmatrix} -\frac{1}{8} + \frac{1}{16} \\ -\frac{1}{8} - \frac{1}{16} \end{bmatrix} = \begin{bmatrix} -1/16 \\ -3/16 \end{bmatrix}\]

    The focus in the \(x'y'\) frame is at \((-\frac{\sqrt{2}}{8}, -\frac{\sqrt{2}}{16} + \frac{\sqrt{2}}{16}) = (-\frac{\sqrt{2}}{8}, 0)\).

    Transform to \(xy\) frame: \[\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{bmatrix} \begin{bmatrix} -\sqrt{2}/8 \\ 0 \end{bmatrix} = \begin{bmatrix} -1/8 \\ -1/8 \end{bmatrix}\]

Answer:

  • Type: Parabola (confirmed)
  • Vertex: \((-\frac{1}{16}, -\frac{3}{16})\)
  • Focus: \((-\frac{1}{8}, -\frac{1}{8})\)
  • Canonical form: \((x'')^2 = \frac{\sqrt{2}}{4}y''\) (in rotated and translated frame)
4.19. Parabola Using Orthogonal Invariants (Lecture 8, Example 9 Alternative)

Use the method of orthogonal invariants to find the canonical form of \(x^2 + 2xy + y^2 + x = 0\).

Click to see the solution

Key Concept: For parabolas, the orthogonal invariant method requires modified formulas.

  1. Identify coefficients:

    \[A = 1, \quad B = 2, \quad C = 1, \quad D = 1, \quad E = 0, \quad F = 0\]

  2. Calculate the determinant:

    \[\Delta = \det\begin{bmatrix} A & B/2 & D/2 \\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{bmatrix} = \det\begin{bmatrix} 1 & 1 & 1/2 \\ 1 & 1 & 0 \\ 1/2 & 0 & 0 \end{bmatrix}\]

    Expanding: \[= 1 \cdot \det\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} - 1 \cdot \det\begin{bmatrix} 1 & 0 \\ 1/2 & 0 \end{bmatrix} + \frac{1}{2} \cdot \det\begin{bmatrix} 1 & 1 \\ 1/2 & 0 \end{bmatrix}\]

    \[= 0 - 0 + \frac{1}{2}(0 - \frac{1}{2}) = -\frac{1}{4}\]

  3. Apply parabola invariant formulas:

    \[\tilde{C} = A + C = 1 + 1 = 2\]

    \[\tilde{D} = 2\sqrt{\frac{-\Delta}{A + C}} = 2\sqrt{\frac{-(-1/4)}{2}} = 2\sqrt{\frac{1/4}{2}} = 2\sqrt{\frac{1}{8}} = 2 \cdot \frac{1}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\]

  4. Write the canonical form:

    \[\tilde{C}y^2 + \tilde{D}x = 0\] \[2y^2 + \frac{\sqrt{2}}{2}x = 0\]

    Solving for the standard parabola form: \[y^2 = -\frac{\sqrt{2}}{4}x\]

    Or equivalently: \[y^2 = 4 \cdot \frac{\sqrt{2}}{16} \cdot x\]

    (Note: The negative sign indicates the parabola opens in the negative \(x\) direction in the canonical frame.)

Answer: Canonical form: \(2y^2 + \frac{\sqrt{2}}{2}x = 0\) or \(y^2 = \frac{\sqrt{2}}{16} \cdot x\) after adjusting for the sign convention.

4.20. Conic to Standard Form (Test 2, Task 1a)

Convert the conic section \(-9x^2 + 16y^2 - 72x - 96y - 144 = 0\) to standard form, identify its type, and calculate its eccentricity.

Click to see the solution

Key Concept: Complete the square for both variables.

  1. Group and factor:

    \[-9x^2 + 16y^2 - 72x - 96y - 144 = 0\] \[-9(x^2 + 8x) + 16(y^2 - 6y) = 144\]

  2. Complete the square:

    \[-9(x^2 + 8x + 16) + 16(y^2 - 6y + 9) = 144 - 144 + 144\] \[-9(x + 4)^2 + 16(y - 3)^2 = 144\]

  3. Divide by 144:

    \[\frac{(y - 3)^2}{9} - \frac{(x + 4)^2}{16} = 1\]

  4. Identify type:

    This is a vertical hyperbola with center \((-4, 3)\), \(a^2 = 9\), \(b^2 = 16\).

  5. Calculate eccentricity:

    For hyperbola: \(c^2 = a^2 + b^2 = 9 + 16 = 25\), so \(c = 5\) \[e = \frac{c}{a} = \frac{5}{3}\]

Answer: Hyperbola \(\dfrac{(y - 3)^2}{9} - \dfrac{(x + 4)^2}{16} = 1\) with eccentricity \(e = \dfrac{5}{3}\)

4.21. Ellipse Properties (Test 2, Task 1b)

For the ellipse \(\dfrac{(x - 2)^2}{9} + \dfrac{(y + 1)^2}{4} = 1\), find its foci, center, vertices, and determine whether it has a horizontal or vertical major axis.

Click to see the solution

Key Concept: Identify \(a\) and \(b\), then use the relationships for ellipses.

  1. Read parameters:

    Center: \((2, -1)\) \(a^2 = 9 \Rightarrow a = 3\) (under \(x\) term, so horizontal) \(b^2 = 4 \Rightarrow b = 2\) (under \(y\) term)

  2. Major axis orientation:

    Since \(a > b\) and \(a\) is under the \(x\) term, the major axis is horizontal.

  3. Find vertices:

    Along major axis: \((2 \pm 3, -1) = (5, -1)\) and \((-1, -1)\) Co-vertices: \((2, -1 \pm 2) = (2, 1)\) and \((2, -3)\)

  4. Find foci:

    \(c^2 = a^2 - b^2 = 9 - 4 = 5 \Rightarrow c = \sqrt{5}\) Foci: \((2 \pm \sqrt{5}, -1)\)

Answer: Center \((2, -1)\), vertices \((5, -1)\) and \((-1, -1)\), foci \((2 \pm \sqrt{5}, -1)\), horizontal major axis

4.22. Parametric Curve as Circle (Test 2, Task 2a)

As \(\theta\) varies from \(0\) to \(2\pi\), the point \(M(2 + 3\cos\theta, 1 + 3\sin\theta)\) traces which curve?

Click to see the solution

Key Concept: Use the Pythagorean identity to eliminate the parameter.

  1. Write parametric equations:

    \(x = 2 + 3\cos\theta\), \(y = 1 + 3\sin\theta\)

  2. Isolate trigonometric functions:

    \(\cos\theta = \dfrac{x - 2}{3}\), \(\sin\theta = \dfrac{y - 1}{3}\)

  3. Apply identity \(\cos^2\theta + \sin^2\theta = 1\):

    \[\left(\frac{x - 2}{3}\right)^2 + \left(\frac{y - 1}{3}\right)^2 = 1\] \[(x - 2)^2 + (y - 1)^2 = 9\]

Answer: Circle with center \((2, 1)\) and radius \(3\)

4.23. Tangent Line to Circle (Test 2, Task 2b)

Find the equation of the tangent line to the circle \(x^2 + y^2 + 4x - 2y - 20 = 0\) at point \(P(2, 4)\).

Click to see the solution

Key Concept: Use implicit differentiation to find the slope.

  1. Define function:

    \(F(x, y) = x^2 + y^2 + 4x - 2y - 20\)

  2. Find partial derivatives:

    \[F_x = 2x + 4\] \[F_y = 2y - 2\]

  3. Calculate slope:

    \[\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{2x + 4}{2y - 2} = -\frac{x + 2}{y - 1}\]

  4. Evaluate at \(P(2, 4)\):

    \[m = -\frac{2 + 2}{4 - 1} = -\frac{4}{3}\]

  5. Write tangent line:

    \[y - 4 = -\frac{4}{3}(x - 2)\] \[3y - 12 = -4x + 8\] \[4x + 3y - 20 = 0\]

Answer: \(4x + 3y - 20 = 0\)

4.24. Conic Type and Rotation Angle (Test 2, Task 3a-i)

Determine the type of conic section \(x^2 - 4xy + y^2 - 12 = 0\) and find the proper angle for rotating axes to remove the \(xy\) term.

Click to see the solution

Key Concept: Use the discriminant and rotation angle formula.

  1. Identify coefficients:

    \(A = 1\), \(B = -4\), \(C = 1\)

  2. Calculate discriminant:

    \[\Delta = B^2 - 4AC = 16 - 4(1)(1) = 12 > 0\]

    Since \(\Delta > 0\), this is a hyperbola.

  3. Find rotation angle:

    \[\tan 2\theta = \frac{B}{A - C} = \frac{-4}{1 - 1} = \infty\]

    So \(2\theta = 90° \Rightarrow \theta = 45°\)

Answer: Hyperbola; rotation angle \(\theta = 45°\)

4.25. Parabola Conic Type (Test 2, Task 3a-ii)

Determine the type of conic section \(x^2 + 2\sqrt{3}xy + 3y^2 + 8\sqrt{3}x - 8y = 0\) and find the proper angle for rotating axes.

Click to see the solution

Key Concept: Use discriminant to identify parabola.

  1. Identify coefficients:

    \(A = 1\), \(B = 2\sqrt{3}\), \(C = 3\)

  2. Calculate discriminant:

    \[\Delta = B^2 - 4AC = (2\sqrt{3})^2 - 4(1)(3) = 12 - 12 = 0\]

    Since \(\Delta = 0\), this is a parabola.

  3. Find rotation angle:

    \[\tan 2\theta = \frac{B}{A - C} = \frac{2\sqrt{3}}{1 - 3} = \frac{2\sqrt{3}}{-2} = -\sqrt{3}\]

    So \(2\theta = 120° \Rightarrow \theta = 60°\)

Answer: Parabola; rotation angle \(\theta = 60°\)

4.26. Distance Ratio Locus (Test 2, Task 3b)

Find all points \(P(x, y)\) such that the distance from \(P\) to \(A(7, 1)\) is twice the distance from \(P\) to \(B(1, 4)\).

Click to see the solution

Key Concept: Set up distance equation and simplify to find the locus.

  1. Write distance condition:

    \[\sqrt{(x - 7)^2 + (y - 1)^2} = 2\sqrt{(x - 1)^2 + (y - 4)^2}\]

  2. Square both sides:

    \[(x - 7)^2 + (y - 1)^2 = 4[(x - 1)^2 + (y - 4)^2]\]

  3. Expand:

    \[x^2 - 14x + 49 + y^2 - 2y + 1 = 4(x^2 - 2x + 1 + y^2 - 8y + 16)\] \[x^2 + y^2 - 14x - 2y + 50 = 4x^2 + 4y^2 - 8x - 32y + 68\]

  4. Rearrange:

    \[3x^2 + 3y^2 + 6x - 30y + 18 = 0\] \[x^2 + y^2 + 2x - 10y + 6 = 0\]

  5. Complete the square:

    \[(x + 1)^2 + (y - 5)^2 = -6 + 1 + 25 = 20\]

Answer: Circle \((x + 1)^2 + (y - 5)^2 = 20\) with center \((-1, 5)\) and radius \(2\sqrt{5}\)

4.27. Circle Through Three Points (Test 2, Task 4a)

Find the equation of a circle passing through points \((1, 2)\), \((2, 4)\), and \((-1, 1)\).

Click to see the solution

Key Concept: Substitute each point into the general circle equation and solve the system.

  1. General form:

    \(x^2 + y^2 + Dx + Ey + F = 0\)

  2. Substitute points:

    For \((1, 2)\): \(1 + 4 + D + 2E + F = 0 \Rightarrow D + 2E + F = -5\) For \((2, 4)\): \(4 + 16 + 2D + 4E + F = 0 \Rightarrow 2D + 4E + F = -20\) For \((-1, 1)\): \(1 + 1 - D + E + F = 0 \Rightarrow -D + E + F = -2\)

  3. Solve system using matrices:

    \[\begin{bmatrix} 1 & 2 & 1 & -5 \\ 2 & 4 & 1 & -20 \\ -1 & 1 & 1 & -2 \end{bmatrix} \xrightarrow{\text{reduce}} D = 3, E = -9, F = 10\]

  4. Write equation:

    \[x^2 + y^2 + 3x - 9y + 10 = 0\]

Answer: \(x^2 + y^2 + 3x - 9y + 10 = 0\)

4.28. Point Classification (Test 2, Task 4b)

Consider the circle \(x^2 + y^2 - 2x + 4y - 4 = 0\) and determine whether points \(A(2, 1)\), \(B(1, 1)\), and \(C(0, 1)\) are inside, outside, or on it.

Click to see the solution

Key Concept: Complete the square and use the power of a point.

  1. Convert to standard form:

    \[x^2 - 2x + y^2 + 4y = 4\] \[(x - 1)^2 + (y + 2)^2 = 4 + 1 + 4 = 9\]

    Center: \((1, -2)\), Radius: \(3\)

  2. Test each point:

    For \(A(2, 1)\): \((2 - 1)^2 + (1 + 2)^2 = 1 + 9 = 10 > 9\) → Outside For \(B(1, 1)\): \((1 - 1)^2 + (1 + 2)^2 = 0 + 9 = 9\) → On the circle For \(C(0, 1)\): \((0 - 1)^2 + (1 + 2)^2 = 1 + 9 = 10 > 9\) → Outside

Answer: \(A\) and \(C\) are outside; \(B\) is on the circle

4.29. Hyperbola Asymptotes (Test 2, Task 4c)

Find the asymptotes of the hyperbola \(\dfrac{y^2}{9} - \dfrac{x^2}{16} = 1\).

Click to see the solution

Key Concept: For vertical hyperbola \(\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1\), asymptotes are \(y = \pm\dfrac{a}{b}x\).

  1. Identify parameters:

    \(a^2 = 9 \Rightarrow a = 3\) \(b^2 = 16 \Rightarrow b = 4\)

  2. Write asymptotes:

    \[y = \pm\frac{3}{4}x\]

Answer: \(y = \dfrac{3}{4}x\) and \(y = -\dfrac{3}{4}x\)

4.30. Convert Mixed-Sign Quadratic to Standard Form (Exercises, Task 1)

Convert the conic section \(-4x^2 + 9y^2 + 24x + 36y - 36 = 0\) to standard form, identify its type, and calculate its eccentricity.

Click to see the solution

Key Concept: Completing the square isolates the quadratic factors and reveals the canonical parameters.

  1. Group the variables and factor common coefficients:

    \[-4(x^2 - 6x) + 9(y^2 + 4y) - 36 = 0\]

  2. Complete the square inside each bracket:

    \[-4\big[(x - 3)^2 - 9\big] + 9\big[(y + 2)^2 - 4\big] - 36 = 0\]

  3. Simplify and divide to reach standard form:

    \[9(y + 2)^2 - 4(x - 3)^2 = 36 \quad \Rightarrow \quad \frac{(y + 2)^2}{4} - \frac{(x - 3)^2}{9} = 1\]

  4. Identify geometry:

    \(a^2 = 4\), \(b^2 = 9\) so

    \[e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{9}{4}} = \frac{\sqrt{13}}{2}\]

Answer: Hyperbola \(\dfrac{(y + 2)^2}{4} - \dfrac{(x - 3)^2}{9} = 1\) with eccentricity \(e = \dfrac{\sqrt{13}}{2}\)

4.31. Analyze a Shifted Ellipse (Exercises, Task 2)

For the ellipse \(\dfrac{(x + 1)^2}{16} + \dfrac{(y - 2)^2}{9} = 1\), find its foci, center, vertices, and determine whether it has a horizontal or vertical major axis.

Click to see the solution
  1. Read parameters directly: Center \((-1, 2)\), \(a^2 = 16\) (\(a = 4\)), \(b^2 = 9\) (\(b = 3\)).
  2. Find vertices: Along the major axis: \((-1 \pm 4, 2)\) gives \((-5, 2)\) and \((3, 2)\). Co-vertices lie \(b\) units along \(y\): \((-1, 2 \pm 3)\).
  3. Compute focal distance: \(c = \sqrt{a^2 - b^2} = \sqrt{16 - 9} = \sqrt{7}\), so foci are \((-1 \pm \sqrt{7}, 2)\).
  4. Determine orientation: The larger denominator sits under \((x + 1)^2\), so the major axis is horizontal.

Answer: Center \((-1, 2)\), vertices \((-5, 2)\) and \((3, 2)\), foci \((-1 \pm \sqrt{7}, 2)\), horizontal major axis

4.32. Identify a Parametric Curve (Exercises, Task 3)

As \(\theta\) varies from \(0\) to \(2\pi\), describe the curve traced by the point \(M(3 + 4\cos\theta, -2 + 4\sin\theta)\) and justify your answer.

Click to see the solution

  1. Isolate the trigonometric functions:

    \[\cos\theta = \frac{x - 3}{4}, \qquad \sin\theta = \frac{y + 2}{4}\]

  2. Apply \(\cos^2\theta + \sin^2\theta = 1\):

    \[\left(\frac{x - 3}{4}\right)^2 + \left(\frac{y + 2}{4}\right)^2 = 1\]

  3. Clear denominators:

    \[(x - 3)^2 + (y + 2)^2 = 16\]

  4. Interpretation: The locus is a circle of radius \(4\) centered at \((3, -2)\).

Answer: Circle \((x - 3)^2 + (y + 2)^2 = 16\) (center \((3, -2)\), radius \(4\))

4.33. Tangent to a Shifted Circle (Exercises, Task 4)

Find the equation of the tangent line to the circle \(x^2 + y^2 - 6x + 4y - 12 = 0\) at the point \(P(1, 3)\).

Click to see the solution

Key Concept: A point of tangency must satisfy the circle; if not, fix the constant term before differentiating.

  1. Complete the square:

    \[(x - 3)^2 + (y + 2)^2 = 25\]

  2. Check the given point:

    For the point \((1, 3)\) to lie on the circle: \((1-3)^2 + (3+2)^2 = 4 + 25 = 29\). The circle equation is \((x - 3)^2 + (y + 2)^2 = 29\).

  3. Implicit differentiation (constant change does not affect derivatives):

    \[2x + 2y \cdot \frac{dy}{dx} - 6 + 4\frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{2x - 6}{2y + 4}\]

  4. Evaluate slope at \(P(1, 3)\):

    \[m = -\frac{2(1) - 6}{2(3) + 4} = -\frac{-4}{10} = \frac{2}{5}\]

  5. Equation via point-slope form:

    \[y - 3 = \frac{2}{5}(x - 1) \quad \Rightarrow \quad y = \frac{2}{5}x + \frac{13}{5}\]

Answer: After correcting the constant to \(-16\), the tangent at \(P(1, 3)\) is \(y = \frac{2}{5}x + \frac{13}{5}\)

4.34. Rotation Classification of a Quadratic (Exercises, Task 5)

For the conic section \(x^2 - 6xy + y^2 - 8 = 0\), determine its type and find the appropriate rotation angle to eliminate the \(xy\) term.

Click to see the solution

  1. Identify coefficients: \(A = 1\), \(B = -6\), \(C = 1\).

  2. Use the discriminant: \(\Delta = B^2 - 4AC = 36 - 4 = 32 > 0\), so the curve is a hyperbola (or a degenerate case).

  3. Compute rotation angle: \(\cot 2\theta = \frac{A - C}{B} = 0 \Rightarrow 2\theta = \frac{\pi}{2}\), hence \(\theta = \frac{\pi}{4}\) (45° rotation).

  4. Apply the rotation via \(u = x + y\), \(v = x - y\) (aligned with 45° axes):

    Substituting gives \(u^2 - 2v^2 + 8 = 0 \Rightarrow \frac{v^2}{4} - \frac{u^2}{8} = 1\), confirming a hyperbola opening along the \(v\)-axis.

Answer: Hyperbola; rotate by \(\theta = 45°\) to obtain \(\dfrac{v^2}{4} - \dfrac{u^2}{8} = 1\)

4.35. Locus with a Distance Ratio (Exercises, Task 6)

Find all points \(P(x, y)\) such that the distance from \(P\) to \(A(3, 2)\) is three times the distance from \(P\) to \(B(-1, 1)\).

Click to see the solution

  1. Translate the ratio into an equation:

    \[\sqrt{(x - 3)^2 + (y - 2)^2} = 3\sqrt{(x + 1)^2 + (y - 1)^2}\]

  2. Square both sides:

    \[(x - 3)^2 + (y - 2)^2 = 9\big[(x + 1)^2 + (y - 1)^2\big]\]

  3. Expand and collect like terms:

    \[8x^2 + 8y^2 + 24x - 14y + 5 = 0\]

  4. Divide by \(8\) and complete the squares:

    \[x^2 + y^2 + 3x - \frac{7}{4}y + \frac{5}{8} = 0\] \[(x + \tfrac{3}{2})^2 + (y - \tfrac{7}{8})^2 = \frac{153}{64}\]

Answer: Apollonius circle centered at \(\left(-\tfrac{3}{2}, \tfrac{7}{8}\right)\) with radius \(\tfrac{\sqrt{153}}{8}\)

4.36. Circle Through Three Points (Exercises, Task 7)

Determine the equation of a circle that passes through the points \((1, 2)\), \((2, 4)\), and \((-2, -1)\).

Click to see the solution

  1. Start from the general form: \(x^2 + y^2 + Dx + Ey + F = 0\).

  2. Substitute each point to build a linear system:

    [

    \[\begin{cases} 1 + 4 + D + 2E + F = 0 \\ 4 + 16 + 2D + 4E + F = 0 \\ 4 + 1 - 2D - E + F = 0 \end{cases} \Rightarrow \begin{cases} D + 2E + F = -5 \\ 2D + 4E + F = -20 \\ -2D - E + F = -5 \end{cases}\]

    ]

  3. Solve: \(D = 15\), \(E = -15\), \(F = 10\).

  4. Write the circle and extract its center/radius:

    \[x^2 + y^2 + 15x - 15y + 10 = 0\] \[(x + 7.5)^2 + (y - 7.5)^2 = \frac{205}{2}\]

Answer: Circle \(x^2 + y^2 + 15x - 15y + 10 = 0\) (center \((-7.5, 7.5)\), radius \(\sqrt{205/2}\))

4.37. Classify Points Relative to a Circle (Exercises, Task 8)

For the circle \(x^2 + y^2 + 4x - 6y - 3 = 0\), determine whether the points \(A(1, 2)\), \(B(-1, 4)\), and \(C(0, 0)\) lie inside, outside, or on the circle.

Click to see the solution

  1. Convert to center-radius form:

    \[(x + 2)^2 + (y - 3)^2 = 16\] Center \((-2, 3)\), radius \(r = 4\).

  2. Use the power test \(P = (x - h)^2 + (y - k)^2 - r^2\):

    • \(A(1, 2)\): \(P_A = 10 - 16 = -6\) (inside)
    • \(B(-1, 4)\): \(P_B = 2 - 16 = -14\) (inside)
    • \(C(0, 0)\): \(P_C = 13 - 16 = -3\) (inside)

Answer: All three points lie inside the circle

4.38. Hyperbola Standard Form and Asymptotes (Exercises, Task 9)

Convert the hyperbola \(9x^2 - 16y^2 - 36x - 64y + 116 = 0\) to standard form and find its asymptotes.

Click to see the solution

  1. Complete the squares:

    \[9(x^2 - 4x) - 16(y^2 + 4y) + 116 = 0\] \[9\big[(x - 2)^2 - 4\big] - 16\big[(y + 2)^2 - 4\big] + 116 = 0\]

  2. Simplify:

    \[9(x - 2)^2 - 16(y + 2)^2 = -144\]

  3. Divide by \(-144\):

    \[\frac{(y + 2)^2}{9} - \frac{(x - 2)^2}{16} = 1\]

  4. State the asymptotes for a vertical hyperbola:

    \[y + 2 = \pm\frac{a}{b}(x - 2) = \pm\frac{3}{4}(x - 2)\]

Answer: Hyperbola \(\dfrac{(y + 2)^2}{9} - \dfrac{(x - 2)^2}{16} = 1\) with asymptotes \(y + 2 = \pm \tfrac{3}{4}(x - 2)\)

4.39. Parabola from Focus and Directrix (Exercises, Task 10)

Find the equation of the parabola with focus at \((2, 3)\) and directrix \(y = -1\).

Click to see the solution

  1. Set equal the squared distances to focus and directrix:

    \[(x - 2)^2 + (y - 3)^2 = (y + 1)^2\]

  2. Expand and simplify:

    \[x^2 - 4x + 4 + y^2 - 6y + 9 = y^2 + 2y + 1\] \[x^2 - 4x - 8y + 12 = 0\]

  3. Complete the square in \(x\):

    \[(x - 2)^2 = 8(y - 1)\]

  4. Interpret parameters: Vertex \((2, 1)\), axis vertical, \(p = 2\) (distance from vertex to focus), focus \((2, 3)\), directrix \(y = -1\).

Answer: Parabola \((x - 2)^2 = 8(y - 1)\) opening upward

4.40. Degenerate Conic via Rotation (Exercises, Task 11)

For the rotated conic \(2x^2 + 4xy + 2y^2 - 36 = 0\), determine the rotation angle needed to eliminate the \(xy\) term and identify the type of conic.

Click to see the solution

  1. Factor the quadratic part:

    \[2(x^2 + 2xy + y^2) - 36 = 2(x + y)^2 - 36 = 0\]

  2. Recognize that \(x + y\) aligns with a 45° axis:

    Since \(A = C\) and \(B \neq 0\), choose \(\theta = 45°\) to remove the mixed term.

  3. Solve the factored equation:

    \[(x + y)^2 = 18 \quad \Rightarrow \quad x + y = \pm 3\sqrt{2}\]

  4. Interpretation: The “conic” is a pair of parallel lines in the rotated frame, i.e., a degenerate case rather than an ellipse/parabola/hyperbola.

Answer: Rotate by \(\theta = 45°\); the equation factors into the parallel lines \(x + y = \pm 3\sqrt{2}\) (degenerate conic)