Precalculus · Chapter 11

Conic Sections

SECTION 11.1

Parabolas

— Geometric Definition of a Parabola

A parabola is the set of all points in the plane that are equidistant from a fixed point \( F \) (called the focus) and a fixed line \( l \) (called the directrix).

— Vertex and Axis of Symmetry

The vertex \( V \) of the parabola lies halfway between the focus and the directrix. The axis of symmetry is the line that runs through the focus perpendicular to the directrix.

— Deriving the Equation

For a parabola with focus \( F(0, p) \) and directrix \( y = -p \), a point \( P(x, y) \) on the curve is equidistant from focus and directrix. Setting the distance to the focus equal to the distance to the directrix, \( \sqrt{x^2 + (y-p)^2} = |y + p| \), and squaring leads directly to \( x^2 = 4py \). The parabola opens upward when \( p > 0 \) and downward when \( p < 0 \).

— Latus Rectum and Focal Diameter

The line segment through the focus, perpendicular to the axis, with endpoints on the parabola is the latus rectum; its length is the focal diameter, equal to \( |4p| \). The focal diameter measures the "width" of the parabola at the focus and is useful for quick sketching.

— Reflection Property

Light from a source at the focus of a parabolic surface reflects into a beam parallel to the axis. Conversely, incoming rays parallel to the axis concentrate at the focus. This property underlies reflecting telescopes, searchlights, and solar collectors.

— Parabola with Vertical Axis

\[ x^2 = 4py \] Vertex \( V(0, 0) \), focus \( F(0, p) \), directrix \( y = -p \). Opens upward if \( p > 0 \), downward if \( p < 0 \).

— Parabola with Horizontal Axis

\[ y^2 = 4px \] Vertex \( V(0, 0) \), focus \( F(p, 0) \), directrix \( x = -p \). Opens right if \( p > 0 \), left if \( p < 0 \).

— Focal Diameter

\[ \text{Focal diameter} = |4p| \]

EXAMPLE 1 Finding the Equation of a Parabola

Find an equation for the parabola with vertex \( V(0, 0) \) and focus \( F(0, 2) \).

Solution

The focus is \( F(0, 2) \), so \( p = 2 \) and the directrix is \( y = -2 \).

Using \( x^2 = 4py \) with \( p = 2 \): \( x^2 = 4(2)y \).

Therefore \( x^2 = 8y \). Since \( p = 2 > 0 \), the parabola opens upward. [!TIP]

EXAMPLE 2 Finding the Focus and Directrix from the Equation

Find the focus and directrix of the parabola \( y = -x^2 \).

Solution

Write in standard form: \( x^2 = -y \).

Compare to \( x^2 = 4py \): \( 4p = -1 \), so \( p = -\tfrac{1}{4} \).

Focus \( F\!\left(0, -\tfrac{1}{4}\right) \); directrix \( y = \tfrac{1}{4} \). [!TIP]

EXAMPLE 3 A Parabola with Horizontal Axis

For \( 6x + y^2 = 0 \), find the focus and directrix.

Solution

Standard form: \( y^2 = -6x \).

Compare to \( y^2 = 4px \): \( 4p = -6 \), so \( p = -\tfrac{3}{2} \).

Focus \( F\!\left(-\tfrac{3}{2}, 0\right) \); directrix \( x = \tfrac{3}{2} \). Since \( p < 0 \), the parabola opens to the left. [!TIP]

EXAMPLE 4 The Focal Diameter of a Parabola

Find the focus, directrix, and focal diameter of \( y = \tfrac{1}{2}x^2 \).

Solution

Rewrite: \( x^2 = 2y \), so \( 4p = 2 \) and the focal diameter is \( 2 \).

Solve for \( p \): \( p = \tfrac{1}{2} \).

Focus \( \left(0, \tfrac{1}{2}\right) \); directrix \( y = -\tfrac{1}{2} \). The latus rectum extends 1 unit each side of the focus. [!TIP]

EXAMPLE 5 A Family of Parabolas

Find equations for parabolas with vertex at the origin and foci \( F_1\!\left(0,\tfrac{1}{8}\right) \), \( F_2\!\left(0,\tfrac{1}{2}\right) \), \( F_3(0,1) \), \( F_4(0,4) \).

Solution

Foci on the positive \( y \)-axis ⇒ form \( x^2 = 4py \).

\( p = \tfrac{1}{8} \Rightarrow x^2 = \tfrac{1}{2}y \); \( p = \tfrac{1}{2} \Rightarrow x^2 = 2y \); \( p = 1 \Rightarrow x^2 = 4y \); \( p = 4 \Rightarrow x^2 = 16y \).

Conclusion: the closer the focus is to the vertex, the narrower the parabola. [!TIP]

EXAMPLE 6 Finding the Focal Point of a Searchlight Reflector

A parabolic reflector bowl is 12 in. wide rim to rim and 8 in. deep. If the bulb filament is at the focus, how far from the vertex is it?

Solution

Place the vertex at the origin, axis vertical: \( x^2 = 4py \). The point \( (6, 8) \) lies on the curve.

Substitute: \( 6^2 = 4p(8) \Rightarrow 36 = 32p \).

Solve: \( p = \tfrac{9}{8} = 1\tfrac{1}{8} \) in. The filament sits \( 1\tfrac{1}{8} \) in. from the vertex.

— A parabola showing focus \( F \), vertex \( V \), directrix \( l \), and axis of symmetry. (Refer to the source course for the figure; reproduced here only by description to respect copyright.)

Figure 11.1.1

— Parabolic reflector concentrating parallel rays at the focus \( F \). (Description only.)

Figure 11.1.2

SECTION 11.2

Ellipses

— Geometric Definition of an Ellipse

An ellipse is the set of all points in the plane the sum of whose distances from two fixed points \( F_1 \) and \( F_2 \) is a constant. The two fixed points are the foci.

— Axes, Vertices, and Center

The vertices are the endpoints of the major axis (length \( 2a \)); the endpoints of the minor axis have length \( 2b \). The major axis is always longer than the minor axis. The center is the midpoint between the foci.

— Relationship Among a, b, c

In standard form, \( a^2 \) is the larger denominator and \( b^2 \) the smaller. The focal distance satisfies \( c^2 = a^2 - b^2 \) (larger denominator minus smaller). If \( a = b \), the ellipse becomes a circle of radius \( a \).

— Eccentricity

Eccentricity \( e = \dfrac{c}{a} \) measures how "stretched" an ellipse is, with \( 0 < e < 1 \). Values near 0 are nearly circular; values near 1 are elongated. Planetary orbits are ellipses with the sun at one focus, most having small eccentricity.

— Reflection Property and Applications

Light or sound from one focus reflects to the other focus. This principle is used in lithotripsy (destroying kidney stones) and in whispering galleries such as the National Statuary Hall.

— Ellipse with Horizontal Major Axis

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \quad a > b > 0 \] Vertices \( (\pm a, 0) \); foci \( (\pm c, 0) \) with \( c^2 = a^2 - b^2 \); major axis length \( 2a \), minor axis length \( 2b \).

— Ellipse with Vertical Major Axis

\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1, \quad a > b > 0 \] Vertices \( (0, \pm a) \); foci \( (0, \pm c) \) with \( c^2 = a^2 - b^2 \).

— Eccentricity

\[ e = \frac{c}{a}, \quad c = \sqrt{a^2 - b^2}, \quad 0 < e < 1 \]

EXAMPLE 1 Sketching an Ellipse

Analyze \( \dfrac{x^2}{9} + \dfrac{y^2}{4} = 1 \): find foci, vertices, and axis lengths.

Solution

Larger denominator under \( x^2 \) ⇒ horizontal major axis; \( a^2 = 9 \), \( b^2 = 4 \).

\( c^2 = a^2 - b^2 = 9 - 4 = 5 \), so \( a = 3 \), \( b = 2 \), \( c = \sqrt{5} \).

Foci \( (\pm\sqrt{5}, 0) \); vertices \( (\pm 3, 0) \); major axis \( 6 \); minor axis \( 4 \). [!TIP]

EXAMPLE 2 Finding the Foci of an Ellipse

Find the foci of \( 16x^2 + 9y^2 = 144 \).

Solution

Divide by 144: \( \dfrac{x^2}{9} + \dfrac{y^2}{16} = 1 \).

Since \( 16 > 9 \), foci are on the \( y \)-axis with \( a = 4 \), \( b = 3 \).

\( c^2 = 16 - 9 = 7 \Rightarrow c = \sqrt{7} \); foci \( (0, \pm\sqrt{7}) \). [!TIP]

EXAMPLE 3 Finding the Equation of an Ellipse

Vertices \( (\pm 4, 0) \), foci \( (\pm 2, 0) \). Find the equation.

Solution

\( a = 4 \) (horizontal major axis), \( c = 2 \).

\( b^2 = a^2 - c^2 = 16 - 4 = 12 \).

Equation: \( \dfrac{x^2}{16} + \dfrac{y^2}{12} = 1 \). [!TIP]

EXAMPLE 4 Finding the Equation from Eccentricity and Foci

Foci \( (0, \pm 8) \), eccentricity \( e = \tfrac{4}{5} \). Find the equation.

Solution

\( e = \tfrac{c}{a} \Rightarrow \tfrac{4}{5} = \tfrac{8}{a} \Rightarrow a = 10 \).

\( b^2 = a^2 - c^2 = 100 - 64 = 36 \Rightarrow b = 6 \).

Foci on the \( y \)-axis ⇒ \( \dfrac{x^2}{36} + \dfrac{y^2}{100} = 1 \).

— Ellipse with foci \( F_1 \), \( F_2 \) and a point \( P \), illustrating the constant‑sum definition. (Description only.)

Figure 11.2.1

SECTION 11.3

Hyperbolas

— Geometric Definition of a Hyperbola

A hyperbola is the set of all points in the plane the difference of whose distances from two fixed points \( F_1 \) and \( F_2 \) is a constant. The two fixed points are the foci. (Compare: an ellipse uses the sum of distances.)

— Branches, Vertices, and Transverse Axis

A hyperbola has two branches. The two vertices lie on the separate branches; the segment joining them is the transverse axis, and its midpoint is the center.

— Asymptotes and the Central Box

A hyperbola approaches two straight asymptotes for large \( x \) and \( y \). To graph: plot \( (\pm a, 0) \) and \( (0, \pm b) \) to form the central box; the box diagonals extend to the asymptotes; then sketch each branch starting at a vertex.

— Reflection Property and Applications

Light aimed at one focus reflects toward the other. This is used in Cassegrain‑type telescopes and in the LORAN navigation system, where intersecting hyperbolas locate a ship from time differences of radio signals.

— How to Sketch a Hyperbola

1. Sketch the central box (crosses axes at \( \pm a \) and \( \pm b \)). 2. Sketch the asymptotes by extending the box diagonals. 3. Plot the vertices (the two intercepts on the transverse axis). 4. Sketch each branch starting at a vertex, approaching the asymptotes.

— Hyperbola with Horizontal Transverse Axis

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] Vertices \( (\pm a, 0) \); foci \( (\pm c, 0) \) with \( c^2 = a^2 + b^2 \); asymptotes \( y = \pm\dfrac{b}{a}x \).

— Hyperbola with Vertical Transverse Axis

\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] Vertices \( (0, \pm a) \); foci \( (0, \pm c) \) with \( c^2 = a^2 + b^2 \); asymptotes \( y = \pm\dfrac{a}{b}x \).

EXAMPLE 1 A Hyperbola with Horizontal Transverse Axis

Analyze \( 9x^2 - 16y^2 = 144 \): find vertices, foci, transverse axis length, and asymptotes.

Solution

Divide by 144: \( \dfrac{x^2}{16} - \dfrac{y^2}{9} = 1 \). Positive \( x^2 \) ⇒ horizontal transverse axis.

\( a^2 = 16 \), \( b^2 = 9 \Rightarrow a = 4 \), \( b = 3 \), \( c = \sqrt{16+9} = 5 \).

Vertices \( (\pm 4, 0) \); foci \( (\pm 5, 0) \); transverse axis \( 2a = 8 \); asymptotes \( y = \pm\tfrac{3}{4}x \). [!TIP]

EXAMPLE 2 A Hyperbola with Vertical Transverse Axis

Analyze \( x^2 - 9y^2 + 9 = 0 \).

Solution

Rearrange and divide by 9: \( y^2 - \dfrac{x^2}{9} = 1 \). Positive \( y^2 \) ⇒ vertical transverse axis.

\( a^2 = 1 \), \( b^2 = 9 \Rightarrow a = 1 \), \( b = 3 \), \( c = \sqrt{1+9} = \sqrt{10} \).

Vertices \( (0, \pm 1) \); foci \( (0, \pm\sqrt{10}) \); transverse axis \( 2 \); asymptotes \( y = \pm\tfrac{1}{3}x \). [!TIP]

EXAMPLE 3 Equation from Vertices and Foci

Vertices \( (\pm 3, 0) \), foci \( (\pm 4, 0) \). Find the equation.

Solution

Horizontal transverse axis, form \( \dfrac{x^2}{3^2} - \dfrac{y^2}{b^2} = 1 \), with \( a = 3 \), \( c = 4 \).

\( a^2 + b^2 = c^2 \Rightarrow 9 + b^2 = 16 \Rightarrow b^2 = 7 \).

Equation: \( \dfrac{x^2}{9} - \dfrac{y^2}{7} = 1 \). [!TIP]

EXAMPLE 4 Equation from Vertices and Asymptotes

Vertices \( (0, \pm 2) \), asymptotes \( y = \pm 2x \). Find the equation and foci.

Solution

Vertical transverse axis ⇒ \( a = 2 \); asymptote slope \( \tfrac{a}{b} = 2 \Rightarrow b = 1 \).

Equation: \( \dfrac{y^2}{4} - x^2 = 1 \).

\( c^2 = a^2 + b^2 = 4 + 1 = 5 \Rightarrow c = \sqrt{5} \); foci \( (0, \pm\sqrt{5}) \).

— Hyperbola with foci \( F_1 \), \( F_2 \): a point \( P \) satisfies \( |d(P,F_1) - d(P,F_2)| = 2a \). (Description only.)

Figure 11.3.1

— Steps in graphing a hyperbola: central box → asymptotes → branches. (Description only.)

Figure 11.3.2

SECTION 11.4

Shifted Conics

— Shifting Graphs

For any equation in \( x \) and \( y \), replacing \( x \) by \( x - h \) shifts the graph right \( h \) units (left for \( x + h \)); replacing \( y \) by \( y - k \) shifts it up \( k \) units (down for \( y + k \)). Conics with vertex/center at \( (h, k) \) use the unshifted form with \( x \to x-h \) and \( y \to y-k \).

— Identifying a Shifted Conic from the General Form

Expanding any shifted conic gives \( Ax^2 + Cy^2 + Dx + Ey + F = 0 \) (with \( A, C \) not both 0). Completing the square recovers the standard form. The signs of \( A \) and \( C \) reveal the type.

— Degenerate Conics

Completing the square may yield a single point, a pair of intersecting lines, or no graph at all (e.g. a sum of squares equal to a negative number). These cases are called degenerate conics.

— Shifted Ellipse

\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] Center \( (h, k) \).

— Shifted Parabola

\[ (x-h)^2 = 4p(y-k) \qquad\text{or}\qquad (y-k)^2 = 4p(x-h) \] Vertex \( (h, k) \).

— Shifted Hyperbola

\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \qquad\text{or}\qquad \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \] Center \( (h, k) \).

— General Equation of a Shifted Conic

\[ Ax^2 + Cy^2 + Dx + Ey + F = 0 \] Nondegenerate cases: **parabola** if \( A = 0 \) or \( C = 0 \); **ellipse** if \( A \) and \( C \) have the same sign (circle if \( A = C \)); **hyperbola** if \( A \) and \( C \) have opposite signs.

EXAMPLE 1 Sketching a Shifted Ellipse

Sketch \( \dfrac{(x+1)^2}{4} + \dfrac{(y-2)^2}{9} = 1 \) and find the foci.

Solution

Center \( (-1, 2) \); obtained from \( \dfrac{x^2}{4} + \dfrac{y^2}{9} = 1 \) shifted left 1, up 2. Here \( a^2 = 9 \), \( b^2 = 4 \) (vertical major axis).

\( c^2 = 9 - 4 = 5 \Rightarrow c = \sqrt{5} \). Unshifted foci \( (0, \pm\sqrt{5}) \).

Shift each: foci \( \left(-1,\, 2+\sqrt{5}\right) \) and \( \left(-1,\, 2-\sqrt{5}\right) \). [!TIP]

EXAMPLE 2 Finding the Equation of a Shifted Ellipse

Vertices \( (-7, 3) \) and \( (3, 3) \); foci \( (-6, 3) \) and \( (2, 3) \). Find the equation.

Solution

Center is the midpoint: \( (-2, 3) \). Horizontal major axis.

Major axis length \( = 10 \Rightarrow a = 5 \); distance between foci \( = 8 \Rightarrow c = 4 \); \( b^2 = 25 - 16 = 9 \).

Equation: \( \dfrac{(x+2)^2}{25} + \dfrac{(y-3)^2}{9} = 1 \). [!TIP]

EXAMPLE 3 Graphing a Shifted Parabola

Analyze \( x^2 - 4x = 8y - 28 \): find vertex, focus, directrix.

Solution

Complete the square: \( x^2 - 4x + 4 = 8y - 28 + 4 \Rightarrow (x-2)^2 = 8y - 24 \).

Factor: \( (x-2)^2 = 8(y-3) \). Vertex \( (2, 3) \); since \( 4p = 8 \), \( p = 2 \).

Focus \( (2, 5) \) (2 units above vertex); directrix \( y = 1 \) (2 units below). [!TIP]

EXAMPLE 4 Graphing a Shifted Hyperbola

Show \( 9x^2 - 72x - 16y^2 - 32y = 16 \) is a hyperbola; find center, vertices, foci, asymptotes.

Solution

Complete the squares: \( 9(x-4)^2 - 16(y+1)^2 = 144 \Rightarrow \dfrac{(x-4)^2}{16} - \dfrac{(y+1)^2}{9} = 1 \).

Center \( (4, -1) \); \( a = 4 \), \( b = 3 \), \( c = \sqrt{16+9} = 5 \).

Foci \( (-1, -1) \) and \( (9, -1) \); vertices \( (0, -1) \) and \( (8, -1) \); asymptotes \( y = \tfrac{3}{4}x - 4 \) and \( y = -\tfrac{3}{4}x + 2 \). [!TIP]

EXAMPLE 5 An Equation Leading to a Degenerate Conic

Sketch \( 9x^2 - y^2 + 18x + 6y = 0 \).

Solution

Complete the squares: \( 9(x+1)^2 - (y-3)^2 = 0 \).

This factors into two lines: \( y - 3 = \pm 3(x+1) \).

Result: a degenerate hyperbola — the pair of intersecting lines \( y = 3x + 6 \) and \( y = -3x \).

— A shifted ellipse with center moved from the origin to \( (h, k) \). (Description only.)

Figure 11.4.1

— The four orientations of shifted parabolas based on the sign of \( p \). (Description only.)

Figure 11.4.2

SECTION 11.5

Rotation of Axes

— Why Rotate Axes

The most general second‑degree equation \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \) includes an \( xy \)-term. Rotating the coordinate axes through an appropriate angle eliminates this term, reducing the equation to a familiar conic in the new \( X\!Y \)-system.

— Choosing the Rotation Angle

To eliminate the \( xy \)-term, rotate through the acute angle \( \phi \) satisfying \( \cot 2\phi = \dfrac{A - C}{B} \). When the value of \( \cot 2\phi \) is not a standard angle, the Half‑Angle Formulas recover \( \cos\phi \) and \( \sin\phi \).

— The Discriminant and Invariance

The quantity \( B^2 - 4AC \), the discriminant, is unchanged (invariant) under any rotation. It identifies the conic type directly, without rotating.

— Rotation of Axes

\[ x = X\cos\phi - Y\sin\phi, \qquad y = X\sin\phi + Y\cos\phi \] \[ X = x\cos\phi + y\sin\phi, \qquad Y = -x\sin\phi + y\cos\phi \]

— Angle That Eliminates the \( xy \)-Term

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

— Identifying Conics by the Discriminant

\[ B^2 - 4AC \] Nondegenerate cases: **parabola** if \( B^2 - 4AC = 0 \); **ellipse** if \( B^2 - 4AC < 0 \); **hyperbola** if \( B^2 - 4AC > 0 \).

EXAMPLE 1 Rotation of Axes

Rotate the axes through \( 30^\circ \) and find the \( X\!Y \)-coordinates of the point with \( xy \)-coordinates \( (2, 4) \).

Solution

\( X = 2\cos 30^\circ + 4\sin 30^\circ = 2\!\left(\tfrac{\sqrt{3}}{2}\right) + 4\!\left(\tfrac{1}{2}\right) = \sqrt{3} + 2 \).

\( Y = -2\sin 30^\circ + 4\cos 30^\circ = -2\!\left(\tfrac{1}{2}\right) + 4\!\left(\tfrac{\sqrt{3}}{2}\right) = -1 + 2\sqrt{3} \).

\( X\!Y \)-coordinates: \( \left(\sqrt{3}+2,\; -1+2\sqrt{3}\right) \). [!TIP]

EXAMPLE 2 Rotating a Hyperbola

Rotate through \( 45^\circ \) to show \( xy = 2 \) is a hyperbola.

Solution

With \( \phi = 45^\circ \): \( x = \tfrac{X}{\sqrt{2}} - \tfrac{Y}{\sqrt{2}} \), \( y = \tfrac{X}{\sqrt{2}} + \tfrac{Y}{\sqrt{2}} \).

Substitute: \( \left(\tfrac{X-Y}{\sqrt{2}}\right)\!\left(\tfrac{X+Y}{\sqrt{2}}\right) = 2 \Rightarrow \tfrac{X^2 - Y^2}{2} = 2 \).

Simplify: \( \dfrac{X^2}{4} - \dfrac{Y^2}{4} = 1 \) — a hyperbola with vertices \( (\pm 2, 0) \) in \( X\!Y \)-coordinates. [!TIP]

EXAMPLE 3 Eliminating the \( xy \)-Term

Eliminate the \( xy \)-term in \( 6\sqrt{3}\,x^2 + 6xy + 4\sqrt{3}\,y^2 = 21\sqrt{3} \) and identify the curve.

Solution

\( \cot 2\phi = \dfrac{6\sqrt{3} - 4\sqrt{3}}{6} = \dfrac{\sqrt{3}}{3} \Rightarrow 2\phi = 60^\circ \Rightarrow \phi = 30^\circ \).

Substitute \( \cos\phi = \tfrac{\sqrt{3}}{2} \), \( \sin\phi = \tfrac{1}{2} \) and simplify to \( 7\sqrt{3}\,X^2 + 3\sqrt{3}\,Y^2 = 21\sqrt{3} \).

Divide: \( \dfrac{X^2}{3} + \dfrac{Y^2}{7} = 1 \) — an ellipse with foci on the \( Y \)-axis. [!TIP]

EXAMPLE 4 Graphing a Rotated Conic

For \( 64x^2 + 96xy + 36y^2 - 15x + 20y - 25 = 0 \), eliminate the \( xy \)-term and identify the graph.

Solution

\( \cot 2\phi = \dfrac{64 - 36}{96} = \dfrac{7}{24} \Rightarrow \cos 2\phi = \tfrac{7}{25} \). Half‑angle: \( \cos\phi = \tfrac{4}{5} \), \( \sin\phi = \tfrac{3}{5} \).

Substituting and simplifying gives \( 100X^2 + 25Y - 25 = 0 \Rightarrow X^2 = -\tfrac{1}{4}(Y - 1) \).

A parabola opening along the negative \( Y \)-axis with vertex \( (0, 1) \) in \( X\!Y \)-coordinates. [!TIP]

EXAMPLE 5 Identifying a Conic by the Discriminant

Identify \( 3x^2 + 5xy - 2y^2 + x - y + 4 = 0 \).

Solution

\( A = 3 \), \( B = 5 \), \( C = -2 \).

\( B^2 - 4AC = 25 - 4(3)(-2) = 25 + 24 = 49 \).

Since \( 49 > 0 \), the conic is a hyperbola.

— Axes rotated through angle \( \phi \), with a point \( P \) given in both \( (x, y) \) and \( (X, Y) \). (Description only.)

Figure 11.5.1

SECTION 11.6

Polar Equations of Conics

— Unified Focus–Directrix Description

All three conics share one description: given a focus \( F \), a directrix \( \ell \), and eccentricity \( e > 0 \), a conic is the set of points \( P \) with \( \dfrac{d(P, F)}{d(P, \ell)} = e \). The conic is a parabola if \( e = 1 \), an ellipse if \( e < 1 \), and a hyperbola if \( e > 1 \).

— Location of the Directrix from the Equation

The form of the polar equation reveals where the directrix lies: \( \cos\theta \) terms correspond to vertical directrices (\( x = \pm d \)); \( \sin\theta \) terms correspond to horizontal directrices (\( y = \pm d \)). The axis of the conic is perpendicular to the directrix.

— Convenience for Rotation

In polar form, rotating a conic about the focus is simple: the graph of \( r = f(\theta - \alpha) \) is the graph of \( r = f(\theta) \) rotated counterclockwise by \( \alpha \). This makes polar equations especially useful for orbits and satellite tracking.

— Polar Equations of Conics

\[ r = \frac{ed}{1 \pm e\cos\theta} \qquad\text{or}\qquad r = \frac{ed}{1 \pm e\sin\theta} \] One focus at the origin, eccentricity \( e \). **Parabola** if \( e = 1 \); **ellipse** if \( 0 < e < 1 \); **hyperbola** if \( e > 1 \).

— Polar Equation of an Ellipse in Terms of \( a \) and \( e \)

\[ r = \frac{a(1 - e^2)}{1 - e\cos\theta} \] Useful for orbital calculations; perihelion distance \( = a(1 - e) \), aphelion distance \( = a(1 + e) \).

EXAMPLE 1 Finding a Polar Equation for a Conic

Find a polar equation for the parabola with focus at the origin and directrix \( y = 6 \).

Solution

Parabola ⇒ \( e = 1 \); directrix \( y = d \) with \( d = 6 \).

Horizontal directrix above the focus ⇒ use \( 1 + \sin\theta \) in the denominator.

Equation: \( r = \dfrac{6}{1 + \sin\theta} \). [!TIP]

EXAMPLE 2 Identifying and Sketching an Ellipse

Analyze \( r = \dfrac{10}{3 - 2\cos\theta} \).

Solution

Divide numerator and denominator by 3: \( r = \dfrac{10/3}{1 - \tfrac{2}{3}\cos\theta} \), so \( e = \tfrac{2}{3} < 1 \) ⇒ ellipse.

Vertices at \( \theta = 0 \) and \( \theta = \pi \): \( V_1(10, 0) \), \( V_2(2, \pi) \); center \( C(4, 0) \); major axis \( 2a = 12 \Rightarrow a = 6 \).

\( c = ae = 6\!\left(\tfrac{2}{3}\right) = 4 \); \( b^2 = a^2 - c^2 = 36 - 16 = 20 \Rightarrow 2b = 4\sqrt{5} \approx 8.94 \). [!TIP]

EXAMPLE 3 Identifying and Sketching a Hyperbola

Analyze \( r = \dfrac{12}{2 + 4\sin\theta} \).

Solution

Divide by 2: \( r = \dfrac{6}{1 + 2\sin\theta} \), so \( e = 2 > 1 \) ⇒ hyperbola.

Vertical transverse axis; vertices \( V_1(2, \pi/2) \), \( V_2(-6, 3\pi/2) \); center \( C(4, \pi/2) \); \( 2a = 4 \Rightarrow a = 2 \).

\( c = ae = 2 \cdot 2 = 4 \); \( b^2 = c^2 - a^2 = 16 - 4 = 12 \Rightarrow b = 2\sqrt{3} \approx 3.46 \), giving the asymptotes via the central box. [!TIP]

EXAMPLE 4 Rotating an Ellipse

Rotate the ellipse of Example 2 through \( \pi/4 \) about the origin.

Solution

Replace \( \theta \) with \( \theta - \tfrac{\pi}{4} \) in the original equation.

New equation: \( r = \dfrac{10}{3 - 2\cos(\theta - \pi/4)} \).

The ellipse is rotated counterclockwise about the focus at the origin.

— Focus at the origin with directrix \( x = d \); a point \( P(r, \theta) \) satisfies \( d(P,F)/d(P,\ell) = e \). (Description only.)

Figure 11.6.1

— Effect of increasing eccentricity: near‑circular ellipse → parabola (\( e=1 \)) → steepening hyperbola. (Description only.)

Figure 11.6.2

— Conics in Buildings

Architects use conic sections for both structural stability and beauty. Examples include circular Roman amphitheaters, the elliptical ceiling of Statuary Hall, the parabolic Skydome roof in Toronto, and hyperbolic/parabolic forms at Washington Dulles Airport and the McDonnell Planetarium. Gaudí used an inverted parabola in La Pedrera, reasoning that a suspended loaded cable forms a parabola and so an inverted parabola best supports a flat roof.

— Constructing a Parabola Geometrically

Using a string of length \( a \) anchored at focus \( F \) and at point \( A \) on a sliding T‑square (also length \( a \)) along a bar \( L \): since \( d(F,P) + d(P,A) = a \) and \( d(L,P) + d(P,A) = a \), it follows that \( d(F,P) = d(L,P) \). The traced curve is therefore a parabola with focus \( F \) and directrix \( L \).

— Curves from Straight Lines

A curved surface can be built from many straight tangent lines (as in the Kobe Tower). For the parabola \( y = x^2 \), the tangent line at \( (a, a^2) \) is \( y = 2ax - a^2 \); the family of all such tangents has the parabola as its envelope.

— A Cut Cylinder Gives an Ellipse

When a cylinder is cut by a plane (e.g. the Tycho Brahe Planetarium), two equal‑radius spheres inscribed in the cylinder touch the plane at \( F_1 \) and \( F_2 \). For any point \( P \) on the cut curve, \( PF_1 + PF_2 \) is constant, proving the curve is an ellipse with foci \( F_1 \) and \( F_2 \).

Instructor Note

Use this space to add experience‑based guidance for this chapter. Suggested placement and starter prompts below — replace the bracketed text with your own notes.

Instructor Note

_[Add your note here — e.g., a memory aid for which way the parabola opens based on the sign of \( p \), or a real shop example of a parabolic reflector.]_

Instructor Note

_[Add your note here — e.g., the "larger denominator under the variable along the major axis" trick, or a worked orbital example.]_

Instructor Note

_[Add your note here — e.g., warn students that for hyperbolas \( c^2 = a^2 + b^2 \), not minus.]_

Instructor Note

_[Add your note here — e.g., a clean checklist for completing the square without sign errors.]_

Instructor Note

_[Add your note here — e.g., when to reach for the discriminant vs. a full rotation.]_

Instructor Note

_[Add your note here — e.g., how to read the directrix location straight off the equation form.]_