Conic Sections

Definition of Conic Sections

The curves that result from the intersection of a plane with a cone are called conic sections. Fig. 40-1 shows the four major possibilities: circle, ellipse, parabola, and hyperbola.

Circle

Ellipse

Parabola

Figure 40-1

Hyperbola

Degenerate cases arise from exceptional situations; for example, if the plane in the first figure that intersects the cone in a circle were to be lowered until it passes through only the vertex of the cone, the circle would “degenerate” into a point. Other degenerate cases are: two intersecting lines, two parallel lines, one line, or no graph at all.

Classification of Second-Degree Equations

The graph of a second-degree equation in two variables $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ is a conic section. Ignoring degenerate cases, the possibilities are as follows:

A. If no xy term is present (B = 0):

  1. If A = C the graph is a circle. Otherwise, $ A C $ ; then:

  2. If AC = 0 the graph is a parabola.

  3. If AC > 0 the graph is an ellipse.

  4. If AC < 0 the graph is a hyperbola.

B. IN GENERAL:

  1. If $ B^{2}-4AC=0 $ the graph is a parabola.

  2. If B2-4AC<0 the graph is an ellipse (or circle if B=0, A=C).

  3. If $ B^{2}-4AC>0 $ the graph is a hyperbola.

The quantity $ B^{2}-4AC $ is called the discriminant of the second-degree equation.

EXAMPLE 40.1 Identify the curve with equation $ x^{2} + 3y^{2} + 8x + 4y = 50 $ , assuming the graph exists. B = 0. Since A = 1 and C = 3, AC = 3 > 0, thus the graph is an ellipse.

EXAMPLE 40.2 Identify the curve with equation $ x^{2} + 8xy + 3y^{2} + 4y = 50 $ , assuming the graph exists. Since A = 1, B = 8, C = 3, $ B^{2} - 4AC = 8^{2} - 4 = 52 > 0 $ , thus the graph is a hyperbola.

SOLVED PROBLEMS

40.1. Derive the classification scheme for second degree equations with B = 0.

First note that any such equation has the form $ Ax^{2} + Cy^{2} + Dx + Ey + F = 0 $ .

  1. If \(A = 0\), the square can be completed on \(y\) to yield \(C(y - k)^{2} = -D(x - h)\); if \(C = 0\), the square can be completed on \(x\) to yield \(A(x - h)^{2} = -E(y - k)\). These are recognizable as equations of parabolas in standard orientation, corresponding to the case \(AC = 0\).

  2. Otherwise neither A nor C is zero. Then the square can be completed on both x and y to yield $ A(x - h)^{2} + C(y - k)^{2} = G $ . The following cases can be further distinguished.

  3. G = 0. The equation represents a degenerate conic section, either a point or two straight lines.

  4. $ G $ . If A and C have opposite signs, the equation can be written as $ - = $ , which is the equation of a hyperbola, corresponding to the case AC < 0. If A and C have both the same sign as G, the equation can be written as $ + = 1 $ . Then if the denominators are equal, this is the equation of a circle; if not, it is the equation of an ellipse, with AC > 0. Finally, if A and C both have the opposite sign from G, the equation represents a degenerate conic section, consisting of no point at all.

40.2. Recall that a rotation of axes through any angle \(\theta\) transforms a second-degree equation of the form \(Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0\) into another second-degree equation in the form \(A'x'^{2} + B'x'y' + C'y'^{2} + D'x' + E'y' + F = 0\). Show that, regardless of the value of \(\theta\), \(B^{2} - 4AC = B'^{2} - 4A'C\).

In Problem 39.2, it was shown that rotating axes through an angle $ $ transforms the equation $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ by making the substitutions $ x = x’ - y’ $ , $ y = x’ + y’ $ , yielding:

\[ \begin{aligned}&x^{\prime2}(A\cos^{2}\theta+B\cos\theta\sin\theta+C\sin^{2}\theta)+x^{\prime}y^{\prime}[-2A\cos\theta\sin\theta+B(\cos^{2}\theta-\sin^{2}\theta)+2C\sin\theta\cos\theta]\\&\quad+y^{\prime2}(A\sin^{2}\theta-B\sin\theta\cos\theta+C\cos^{2}\theta)+x^{\prime}(D\cos\theta+E\sin\theta)+y^{\prime}(-D\sin\theta+E\cos\theta)+F=0\\ \end{aligned} \]

Comparing this with the form $ A{}x{} + B{}x{}y^{} + C{}y{} + D{}x{} + E{}y{} + F = 0 $ shows that

\[ A^{\prime}=A\cos^{2}\theta+B\cos\theta\sin\theta+C\sin^{2}\theta \]

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

\[ C^{\prime}=A\sin^{2}\theta-B\sin\theta\cos\theta+C\cos^{2}\theta \]

Thus

\[ \begin{aligned}B^{\prime2}-4A^{\prime}C^{\prime}&=[-2A\cos\theta\sin\theta+B(\cos^{2}\theta-\sin^{2}\theta)+2C\sin\theta\cos\theta]^{2}\\&-4(A\cos^{2}\theta+B\cos\theta\sin\theta+C\sin^{2}\theta)(A\sin^{2}\theta-B\sin\theta\cos\theta+C\cos^{2}\theta)\end{aligned} \]

Expanding and collecting terms yields

\[ \begin{aligned}B^{\prime2}-&4A^{\prime}C^{\prime}=A^{2}(4\cos^{2}\theta\sin^{2}\theta-4\cos^{2}\theta\sin^{2}\theta)+B^{2}(\cos^{4}\theta-2\cos^{2}\theta\sin^{2}\theta+\sin^{4}\theta+4\cos^{2}\theta\sin^{2}\theta)\\&+C^{2}(4\cos^{2}\theta\sin^{2}\theta-4\cos^{2}\theta\sin^{2}\theta)+AB(-4\cos^{3}\theta\sin\theta+4\cos\theta\sin^{3}\theta+4\cos^{3}\theta\sin\theta-4\cos\theta\sin^{3}\theta)\\&+AC(-8\sin^{2}\theta\cos^{2}\theta-4\cos^{4}\theta-4\sin^{4}\theta)+BC(4\cos^{3}\theta\sin\theta-4\cos\theta\sin^{3}\theta-4\cos^{3}\theta\sin\theta+4\cos\theta\sin^{3}\theta)\end{aligned} \]

The coefficients of $ A^{2} $ , $ C^{2} $ , AB, and BC are seen to reduce to zero, and the right side reduces to:

\[ \begin{aligned}B^{\prime2}-4A^{\prime}C^{\prime}&=B^{2}(\cos^{4}\theta+2\cos^{2}\theta\sin^{2}\theta+\sin^{4}\theta)-4AC(\cos^{4}\theta+2\cos^{2}\theta\sin^{2}\theta+\sin^{4}\theta)\\&=(B^{2}-4AC)(\cos^{2}\theta+\sin^{2}\theta)^{2}\\&=B^{2}-4AC\\ \end{aligned} \]

This equality is often stated as follows: The quantity $ B^{2}-4AC $ is invariant under a rotation of axes through any angle.

40.3. Derive the general classification scheme for second-degree equations.

The general second-degree equation has the form $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ . If $ B $ , then it was shown in the previous chapter (Problem 39.2) that there is an angle $ $ through which the axes can be rotated so that the equation takes the form $ A’x’^{2} + C’y’^{2} + D’x’ + E’y’ + F = 0 $ . Then, this is the equation of

  1. A parabola if $ A’C’ = 0 $

  2. An ellipse if $ A’C’ > 0 $

  3. A hyperbola if $ A{}C{}<0 $

For the transformed equation, the discriminant becomes $ B^{2} - 4AC = -4A’C’ $ . Therefore, in case 1, the original discriminant $ B^{2} - 4AC = 0 $ ; in case 2, $ B^{2} - 4AC < 0 $ ; and in case 3, $ B^{2} - 4AC > 0 $ . Summarizing, the equation $ Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0 $ represents

  1. A parabola if $ B^{2}-4AC=0 $

  2. An ellipse if $ B^{2}-4AC<0 $ (or circle if B=0, A=C)

  3. A hyperbola if $ B^{2} - 4AC > 0 $

Here degenerate cases are neglected and it is assumed that the equation has a graph.

40.4. Identify the following as the equations of a circle, an ellipse, a parabola, or a hyperbola:

  1. $ 3x^{2} + 8x + 12y = 16 $ ; (b) $ 3x^{2} - 3y^{2} + 8x + 12y = 16 $ ;

  2. $ 3x^{2} + 3y^{2} + 8x + 12y = 16 $ ; (d) $ 3x^{2} + 4y^{2} + 8x + 12y = 16 $

  3. Here B = 0, A = 3, and C = 0. With B = 0, since AC = 0, this is the equation of a parabola.

  4. Here B = 0, A = 3, and C = -3. With B = 0, since AC < 0, this is the equation of a hyperbola.

  5. Here B = 0, and A = C = 3. Thus this is the equation of a circle.

  6. Here B = 0, A = 3, and C = 4. With B = 0, since AC > 0, this is the equation of an ellipse

40.5. Identify the following as the equations of a circle, an ellipse, a parabola, or a hyperbola:

  1. $ 3x^{2} + 8xy + 12y = 16 $ ; (b) $ 3x^{2} + 8xy - 3y^{2} + 8x + 12y = 16 $ ;

\[ \begin{aligned}(c)3x^{2}+6xy+3y^{2}+8x+12y&=16;(d)3x^{2}+2xy+3y^{2}+8x+12y&=16\end{aligned} \]

  1. Here \(A = 3, B = 8, C = 0\), so \(B^{2} - 4AC = 8^{2} - 4 \cdot 3 \cdot 0 = 64 > 0\). Hence this is the equation of a hyperbola.

  2. Here A = 3, B = 8, C = -3, so $ B^{2} - 4AC = 8^{2} - 4 (-3) = 100 > 0 $ . Hence this is the equation of a hyperbola.

  3. Here A = 3, B = 6, C = 3, so $ B^{2} - 4AC = 6^{2} - 4 = 0 $ . Hence this is the equation of a parabola.

  4. Here A = 3, B = 2, C = 3, so $ B^{2} - 4AC = 2^{2} - 4 = -32 < 0 $ . Hence this is the equation of an ellipse.

Note: Since \(B \neq 0\) in all of these cases, none of these can be an equation of a circle.

40.6. It can be shown that, in general, for any ellipse and any hyperbola, there are two straight lines called directrices, perpendicular to the focal axis and at distance $ a/e = a^{2}/c $ from the center, such that the equation of the curve can be derived from the relation $ PF = e PD $ , where PF is the distance from a point on the curve to a focus and PD is the perpendicular distance to the directrix. (See Figs. 40-2 and 40-3.)

Find the directives and verify the derivation of the equation from $ PF = e PD $ for:

  1. the ellipse $ + y^{2} = 1 $ ; (b) the hyperbola $ x^{2} - y^{2} = 1 $

  2. Here a = 2, b = 1, $ c = = = $ . Thus $ e = c/a = /2 $ . The directives then are the vertical lines $ x = a/e = /2 = / $ . The relation $ PF = e PD $ then becomes

\[ \sqrt{(x-\sqrt{3})^{2}+y^{2}}=\frac{\sqrt{3}}{2}\bigg|x-\frac{4}{\sqrt{3}} \]

(choosing the right-hand focus and directrix)

Squaring both sides and simplifying yields:

\[ \begin{aligned}x^{2}-2x\sqrt{3}+3+y^{2}&=\frac{3}{4}\bigg(x^{2}-\frac{8}{\sqrt{3}}x+\frac{16}{3}\bigg)\\x^{2}-2x\sqrt{3}+3+y^{2}&=\frac{3}{4}x^{2}-2x\sqrt{3}+4\\\frac{x^{2}}{4}+y^{2}&=1\end{aligned} \]

  1. Here a = 1, b = 1, $ c = = = $ . Thus $ e = c/a = $ . The directives then are the vertical lines $ x = a/e = / $ . The relation $ PF = e PD $ then becomes

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

(choosing the right-hand focus and directrix)

Squaring both sides and simplifying yields:

\[ \begin{aligned}x^{2}-2x\sqrt{2}+2+y^{2}&=2\left(x^{2}-\frac{2}{\sqrt{2}}x+\frac{1}{2}\right)\\x^{2}-2x\sqrt{2}+2+y^{2}&=2x^{2}-2x\sqrt{2}+1\\x^{2}+y^{2}+2&=2x^{2}+1\\1&=x^{2}-y^{2}\end{aligned} \]

Note: If the eccentricity of a parabola is defined as 1, then the relation $ PF = e PD $ can be viewed as describing all three noncircle conic sections: parabola, ellipse, and hyperbola.

40.7. Show that in polar coordinates the equation of a conic section with (one) focus at the pole and directrix, the line $ r = -p $ can be written as

\[ r=\frac{ep}{1-e\cos\theta} \]

See Fig. 40-4.

Figure 40-4

From the previous problem, a conic section can be defined by the relation $ PF = e PD $ . Here, the focus is at the origin, hence the distance from a point on the conic section to the focus PF = r. The distance from P to the directrix is given by

\[ PD=PA+AD=r\cos\theta+p. \]

Thus

\[ \begin{aligned}PF&=e\cdot PD\\r&=e(r\cos\theta+p)\\r&=re\cos\theta+ep\\r-re\cos\theta&=ep\\r(1-e\cos\theta)&=ep\\r&=\frac{ep}{1-e\cos\theta}\end{aligned} \]

40.8. Identify each of the following as the equations of an ellipse, a hyperbola, or a parabola:

  1. Comparing the given equation with $ r = $ , e = 1 and ep = p = 4, hence this is the equation of a parabola.

  2. Comparing the given equation with $ r = $ , e = 2 and ep = 2p = 4, p = 2, hence this is the equation of a hyperbola.

  3. To compare the given equation with $ r = $ , rewrite it as $ r = $ . Then $ e = $ and $ ep = p = 2 $ , p = 4, hence this is the equation of an ellipse.

SUPPLEMENTARY PROBLEMS

40.9. Identify the following as the equations of a circle, an ellipse, a parabola, or a hyperbola:

\[ \begin{aligned}(a)x^{2}+2y^{2}-2x+3y&=50;(b)x^{2}-2x+3y=50;(c)x^{2}-y^{2}-2x+3y=50;\end{aligned} \]

\[ \left(\mathrm{d}\right)y^{2}-2x+3y=x^{2}+50;\left(\mathrm{e}\right)2x^{2}+2y^{2}-2x+3y=50 \]

Ans. (a) ellipse; (b) parabola; (c) hyperbola; (d) hyperbola; (e) circle

40.10. Identify the following as the equations of a circle, an ellipse, a parabola, or a hyperbola:

\[ (a)x^{2}+2xy+y^{2}-2x+3y=50;(b)2xy+y^{2}-2x+3y=50;(c)x^{2}+xy+y^{2}-2x+3y=50; \]

\[ \left(\mathrm{d}\right)x^{2}+4xy+y^{2}-2x+3y=50;\left(\mathrm{e}\right)(x-y)^{2}+(x+y)^{2}-2x=50 \]

Ans. (a) parabola; (b) hyperbola; (c) ellipse; (d) hyperbola; (e) circle

40.11. The following equations represent typical degenerate conic sections. By factoring or other algebraic techniques, identify the graphs:

\[ (a)x^{2}+xy-3x=0;(b)x^{2}-2xy+y^{2}=81;(c)x^{2}+4xy+4y^{2}+2x+4y+1=0; \]

\[ \left(\mathrm{d}\right)2x^{2}+y^{2}-4y+16=0;\left(\mathrm{e}\right)2x^{2}+4x+y^{2}-4y+6=0 \]

Ans. (a) x = 0 or $ x + y = 3 $ , two intersecting lines; (b) $ x - y = $ , two parallel lines;

  1. $ x + 2y + 1 = 0 $ , one line; (d) $ 2x^{2} + (y - 2)^{2} = -12 $ , no point;

  2. $ 2(x+1){2}+(y-2){2}=0 $ , graph contains one point: $ (-1,2) $

40.12. Find the directives for the graphs of the following equations: (a) $ + = 1 $ ; (b) $ - = 1 $

\[ Ans.\quad(a)\quad y=\pm\frac{9}{\sqrt{5}};(b)\ y=\pm\frac{9}{\sqrt{13}} \]

40.13. Show that in polar coordinates the equation of a conic section with (one) focus at the pole and directrix the line $ r = p $ can be written as

\[ r=\frac{ep}{1+e\cos\theta} \]

40.14. Identify each of the following as the equations of an ellipse, a hyperbola, or a parabola:

\[ \begin{aligned}(a)r=\frac{12}{3-\cos\theta};(b)r=\frac{12}{1+3\cos\theta};(c)r=\frac{12}{1+\cos\theta};(d)r=\frac{12}{3-8\cos\theta}\end{aligned} \]

Ans. (a) ellipse; (b) hyperbola; (c) parabola; (d) hyperbola

40.15. Show that in polar coordinates the equation of a conic section with (one) focus at the pole and directrix the line $ r = p $ can be written as

\[ r=\frac{ep}{1+e\sin\theta} \]

40.16. Show that in polar coordinates the equation of a conic section with (one) focus at the pole and directrix the line $ r = -p $ can be written as

\[ r=\frac{ep}{1-e\sin\theta} \]