Conic sections: parabolas, ellipses and hyperbolas

Worked solutions

Problem 1

1. $4p = 12$, so $p = 3$Match $y^2 = 4px$.
2. Focus $(3, 0)$, directrix $x = -3$$(p, 0)$ and $x = -p$.
3. At $x = 3$: $y^2 = 36$, so $y = \pm 6$The chord through the focus perpendicular to the $x$-axis.
4. $(3, 6)$: distance to the focus $6$, to the directrix $3 + 3 = 6$The focus is directly below the point; the directrix is a vertical line.
5. Focus $(3, 0)$, directrix $x = -3$, latus rectum $12 = 4p$The defining property holds at every point of the curve, which is what makes a parabolic mirror send every ray parallel to the axis through the focus.

Problem 2

1. $\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1$Divide by $225$.
2. $a = 5$, $b = 3$The square roots of the denominators; the larger goes with $x$, so the major axis is horizontal.
3. $c = \sqrt{25 - 9} = 4$For an ellipse $c^2 = a^2 - b^2$.
4. $\sqrt{7^2 + 2.4^2} + \sqrt{1^2 + 2.4^2} = 7.4 + 2.6$Distances from $(3, 2.4)$ to $(-4, 0)$ and to $(4, 0)$.
5. $a = 5$, $b = 3$, $c = 4$; vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$, $e = 0.8$; the distances add up to $10 = 2a$The point is on the ellipse: $9 \cdot 9 + 25 \cdot 5.76 = 225$. Using $c^2 = a^2 + b^2$ instead puts the foci outside the ellipse (Mistake 1).

Problem 3

1. $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$Divide by $144$.
2. $a = 3$, $b = 4$The square roots of $9$ and $16$; the positive term is in $x$, so it opens left and right.
3. $c = \sqrt{9 + 16} = 5$For a hyperbola $c^2 = a^2 + b^2$.
4. Asymptotes $y = \pm\tfrac43 x$$\pm b/a$: for large $x$ the $1$ on the right is negligible, so $16x^2 \approx 9y^2$.
5. Vertices $(\pm 3, 0)$, foci $(\pm 5, 0)$, asymptotes $y = \pm\tfrac43 x$, $e = \tfrac53$Check at $(5, \tfrac{16}{3})$, which is on the curve: its distances to the foci are $\tfrac{34}{3}$ and $\tfrac{16}{3}$, differing by $6 = 2a$.

Problem 4

1. $(x^2 - 6x) + 4(y^2 + 4y) = -21$Group the $x$ terms and the $y$ terms; factor the $4$ out of the $y$ terms.
2. $(x - 3)^2 - 9 + 4\big((y + 2)^2 - 4\big) = -21$Complete each square inside its own parentheses.
3. $(x - 3)^2 + 4(y + 2)^2 = -21 + 9 + 16 = 4$The $-4$ inside the parentheses is multiplied by $4$, so $16$ moves to the right.
4. $\dfrac{(x - 3)^2}{4} + \dfrac{(y + 2)^2}{1} = 1$Divide by $4$.
5. Center $(3, -2)$, $a = 2$, $b = 1$, $c = \sqrt3$; foci $(3 \pm \sqrt3, -2)$A horizontal ellipse. Moving only $4$ instead of $16$ to the right gives $-8$, an equation with no graph at all (Mistake 3).

Problem 5

1. Ellipse: $2a = 10$, so $a = 5$; $c = 3$The major axis runs between the vertices.
2. $b^2 = a^2 - c^2 = 25 - 9 = 16$From $c^2 = a^2 - b^2$.
3. Hyperbola: vertices on the $y$-axis, so $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ with $a = 2$The positive term goes with the axis the vertices are on.
4. Asymptotes $y = \pm\tfrac{a}{b}x$, so $\tfrac{2}{b} = \tfrac12$ and $b = 4$For a vertical hyperbola the slope is $a/b$.
5. Ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$; hyperbola $\dfrac{y^2}{4} - \dfrac{x^2}{16} = 1$Check the asymptotes: for large $\lvert x\rvert$, $y^2 \approx 4 \cdot x^2/16$, so $y \approx \pm x/2$. Taking the slope as $b/a$ gives $b = 1$ and asymptotes $y = \pm 2x$ (Mistake 2).

Problem 6

1. (a) $B^2 - 4AC = 1 - 4 = -3 < 0$$A = 1$, $B = 1$, $C = 1$.
2. (b) $B^2 - 4AC = 16 - 4 = 12 > 0$$A = 1$, $B = 4$, $C = 1$.
3. (c) $B^2 - 4AC = 4 - 4 = 0$$A = 1$, $B = 2$, $C = 1$.
4. (c) $(x + y)^2 = -x$The quadratic part is a perfect square, the mark of a parabola.
5. (a) ellipse, (b) hyperbola, (c) parabolaIn (b) both squared terms are positive, but the $xy$ term outweighs them: along $y = -x$ the left side is $-2x^2$, which takes the curve's equation negative (Mistake 5).

Problem 7

1. $\cot 2\theta = (A - C)/B = 0$, so $\theta = 45^\circ$$A = C$.
2. $x = \tfrac{u - v}{\sqrt2}$, $y = \tfrac{u + v}{\sqrt2}$The rotation by $45^\circ$.
3. $x^2 + y^2 = u^2 + v^2$ and $xy = \tfrac12(u^2 - v^2)$Expand both; the $uv$ terms cancel.
4. $\tfrac32 u^2 + \tfrac12 v^2 = 3$, that is $\dfrac{u^2}{2} + \dfrac{v^2}{6} = 1$Substitute and divide by $3$.
5. Semi-axes $\sqrt6$ along $y = -x$ (the $v$ direction) and $\sqrt2$ along $y = x$Check: on $y = -x$, $x^2 = 3$, so $x = \sqrt3$ and the distance is $\sqrt6$; on $y = x$, $3x^2 = 3$ and the distance is $\sqrt2$. The coefficients $\tfrac32$ and $\tfrac12$ are the eigenvalues of $\begin{bmatrix} 1 & 1/2 \\ 1/2 & 1 \end{bmatrix}$.

Problem 8

1. $\tfrac{2x}{25} + \tfrac{2y}{16}\,y' = 0$Differentiate both sides in $x$; $y$ is a function of $x$.
2. $y' = -\dfrac{16x}{25y}$Solve for $y'$.
3. At $(3, \tfrac{16}{5})$: $y' = -\dfrac{48}{80} = -\tfrac35$$25 \cdot \tfrac{16}{5} = 80$.
4. $y - \tfrac{16}{5} = -\tfrac35(x - 3)$Point-slope form.
5. $y = 5 - \tfrac35 x$, the same as $\tfrac{3x}{25} + \tfrac{y}{5} = 1$The general tangent to an ellipse at $(x_0, y_0)$ is $\tfrac{x x_0}{a^2} + \tfrac{y y_0}{b^2} = 1$.

Problem 9

1. $r = \dfrac{3}{1 + \tfrac12\cos\theta}$Divide the top and bottom by $2$ to get the form $ed/(1 + e\cos\theta)$.
2. $e = \tfrac12$, so an ellipse; $ed = 3$ gives $d = 6$Directrix $x = 6$.
3. $r(0) = 2$ and $r(\pi) = 6$The vertices are $(2, 0)$ and $(-6, 0)$.
4. $2a = 8$, center $(-2, 0)$, $c = 2$, $b^2 = 16 - 4 = 12$The center is halfway between the vertices, and the focus at the origin is $c = 2$ from it.
5. An ellipse with $e = \tfrac12$, directrix $x = 6$: $\dfrac{(x + 2)^2}{16} + \dfrac{y^2}{12} = 1$Check: $c/a = 2/4 = e$. This is the shape of a planet's orbit, with the sun at the focus at the origin.

Problem 10

1. Upper half: $y = 3\sqrt{1 - x^2/25}$Solve $\tfrac{x^2}{25} + \tfrac{y^2}{9} = 1$ for $y \ge 0$.
2. Area $= 4\int_0^5 3\sqrt{1 - x^2/25}\,dx$Four equal quarters, by symmetry.
3. $x = 5\sin t$, $dx = 5\cos t\,dt$Trig substitution; $t$ runs from $0$ to $\tfrac{\pi}{2}$.
4. $= 60\int_0^{\pi/2}\cos^2 t\,dt = 60 \cdot \tfrac{\pi}{4}$$\sqrt{1 - \sin^2 t} = \cos t$ on this interval.
5. Area $= 15\pi \approx 47.12 = \pi \cdot 5 \cdot 3$In general the ellipse is the unit circle stretched by $a$ in $x$ and $b$ in $y$, which multiplies its area $\pi$ by $ab$.