Practice questions
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Chapter 7: Conic sections and parametric equations
Try each problem first. Then check your work with the solution.
1. Parabolas.
(a) Find the focus and directrix of ${y^2=-20x}$.
(b) Find the equation of the parabola with vertex at the origin and directrix ${y=3}$.
(c) Find the vertex, focus, and directrix of ${x^2+6x-4y+1=0}$.
Solution:
(a)
${4p=-20}$, so ${p=-5}$. The focus is ${(-5,0)}$ and the directrix is ${x=5}$. The parabola opens to the left.
(b)
The directrix ${y=3}$ is above the vertex, so the focus is below, at ${(0,-3)}$, and ${p=-3}$:
${x^2=4(-3)y=-12y}$
(c)
Complete the square in ${x}$. Half of ${6}$ is ${3}$, and ${3^2=9}$:
$\begin{align*}x^2+6x&=4y-1\\x^2+6x+9&=4y+8\\(x+3)^2&=4(y+2)\end{align*}$
The vertex is ${(-3,-2)}$ and ${4p=4}$, so ${p=1}$. It opens up. The focus is ${(-3,-1)}$, and the directrix is ${y=-3}$.
2. Ellipses.
(a) Find the vertices, foci, and eccentricity of ${\dfrac{x^2}{9}+\dfrac{y^2}{25}=1}$.
(b) Find the equation of the ellipse with foci ${(\pm 3,0)}$ and vertices ${(\pm 5,0)}$.
Solution:
(a)
The larger denominator is under ${y^2}$, so the ellipse is tall, with ${a=5}$ and ${b=3}$. Then ${c^2=25-9=16}$, so ${c=4}$.
Vertices ${(0,\pm 5)}$, foci ${(0,\pm 4)}$, eccentricity ${e=\dfrac{4}{5}=0.8}$.
(b)
${a=5}$ and ${c=3}$, so ${b^2=25-9=16}$:
${\dfrac{x^2}{25}+\dfrac{y^2}{16}=1}$
3. Hyperbolas.
(a) Find the vertices, foci, and asymptotes of ${9x^2-4y^2=36}$.
(b) Find the equation of the hyperbola with vertices ${(\pm 3,0)}$ and asymptotes ${y=\pm 2x}$.
Solution:
(a)
Divide by ${36}$:
${\dfrac{x^2}{4}-\dfrac{y^2}{9}=1}$
So ${a=2}$, ${b=3}$, and ${c=\sqrt{4+9}=\sqrt{13}}$. It opens left and right.
Vertices ${(\pm 2,0)}$, foci ${(\pm\sqrt{13},0)}$, asymptotes ${y=\pm\dfrac{3}{2}x}$.
(b)
${a=3}$, and the asymptote slope is ${\dfrac{b}{a}=2}$, so ${b=6}$:
${\dfrac{x^2}{9}-\dfrac{y^2}{36}=1}$
4. Rotation of axes.
(a) Identify the conic ${x^2-2xy+y^2+x=0}$.
(b) Identify the conic ${2x^2+3xy+2y^2=7}$.
(c) Rotate the axes to remove the ${xy}$ term from ${xy=-8}$, and identify the graph.
Solution:
(a)
${A=1}$, ${B=-2}$, ${C=1}$, so ${B^2-4AC=4-4=0}$. A parabola.
(b)
${B^2-4AC=9-16=-7<0}$. An ellipse.
(c)
${A=C=0}$, so ${\varphi=45^\circ}$, with ${x=\dfrac{X-Y}{\sqrt{2}}}$ and ${y=\dfrac{X+Y}{\sqrt{2}}}$. Then ${xy=\dfrac{X^2-Y^2}{2}}$, so
$\begin{align*}\dfrac{X^2-Y^2}{2}&=-8\\\dfrac{Y^2}{16}-\dfrac{X^2}{16}&=1\end{align*}$
A hyperbola opening along the ${Y}$-axis, which is the line ${y=-x}$.
5. Polar equations of conics.
(a) Identify ${r=\dfrac{4}{1-\sin\theta}}$, and find its directrix.
(b) Identify ${r=\dfrac{15}{5+3\cos\theta}}$, and find its vertices.
Solution:
(a)
The denominator starts with ${1}$, and ${e=1}$: a parabola. From ${ed=4}$, ${d=4}$. The term ${-\sin\theta}$ puts the directrix below: ${y=-4}$.
(b)
Divide the top and bottom by ${5}$:
${r=\dfrac{3}{1+\dfrac{3}{5}\cos\theta}}$
So ${e=\dfrac{3}{5}<1}$: an ellipse.
At ${\theta=0}$: ${r=\dfrac{15}{8}}$, the vertex ${\left(\dfrac{15}{8},0\right)}$.
At ${\theta=\pi}$: ${r=\dfrac{15}{2}}$, which is ${7.5}$ units in the direction ${\pi}$: the vertex ${\left(-\dfrac{15}{2},0\right)}$.
6. Parametric equations.
(a) Eliminate the parameter: ${x=2t-1}$, ${y=t^2}$.
(b) Eliminate the parameter, and describe the curve: ${x=2+\cos t}$, ${y=-1+\sin t}$, ${0\le t\le 2\pi}$.
(c) A ball is thrown at ${80}$ ft/s at ${30^\circ}$ above level ground. Find when and where it lands, and its greatest height.
Solution:
(a)
Solve for ${t}$: ${t=\dfrac{x+1}{2}}$. Substitute:
$y=\left(\dfrac{x+1}{2}\right)^2=\dfrac{(x+1)^2}{4}$
A parabola with vertex ${(-1,0)}$.
(b)
${\cos t=x-2}$ and ${\sin t=y+1}$. Since ${\cos^2 t+\sin^2 t=1}$,
${(x-2)^2+(y+1)^2=1}$
The circle with center ${(2,-1)}$ and radius ${1}$, traced once counterclockwise.
(c)
${80\cos 30^\circ=40\sqrt{3}}$ and ${80\sin 30^\circ=40}$, so
${x=40\sqrt{3}\,t}$${y=40t-16t^2}$
Landing: ${y=8t(5-2t)=0}$, so ${t=2.5}$ s. Then
$\begin{align*}x&=40\sqrt{3}\cdot 2.5\\&=100\sqrt{3}\approx 173.2\end{align*}$
It lands about ${173.2}$ ft away.
Greatest height: halfway through the flight, at ${t=1.25}$:
$\begin{align*}y&=40(1.25)-16(1.25)^2\\&=50-25=25\end{align*}$
The greatest height is ${25}$ ft.