## Exercises4.4Conic Sections

Solve the following problems.

###### 1.

Sketch the graph of the ellipse $\ds \frac{x^2}{9}+\frac{y^2}{16}=1$ and determine its foci.

Focci: $(0,-\sqrt{7})\text{,}$ $(0,\sqrt{7})\text{.}$ ###### 2.

Let $C$ be the conic which consists of all points $P=(x,y)$ such that the distance from $P$ to $F=(0,1)$ is $\ds \frac{1}{2}$ the perpendicular distance from $P$ to the line $y=4\text{.}$

1. What is the eccentricity of the conic $C\text{?}$

2. Show that the equation of the conic $C$ is

\begin{equation*} \frac{x^2}{3}+\frac{y^2}{4}=1\text{.} \end{equation*}
3. What is the equation of the tangent line to the point $A=(\sqrt{3}/2,\sqrt{3})$ on the conic $C\text{?}$

1. $\ds e=\frac{1}{2}\text{.}$

2. Use the fact that, for $P=(x,y)\text{,}$ $\ds |PF|^2=x^2+(y-1)^2$ and $\ds |Pl|=\frac{1}{2}|y-4|\text{.}$

3. From $\ds \frac{dy}{dx}=-\frac{4x}{3y}$ it follows that the slope of the tangent line is $\ds \left. \frac{dy}{dx}\right| _{x=\frac{\sqrt{3}}{2}}=-\frac{2}{3}\text{.}$

###### 3.
1. Find the equation of the tangent line to the polar curve $r=1+\cos \theta$ at $\ds \theta =\frac{\pi }{2}\text{.}$

2. Find an equation of the ellipse with horizontal and vertical axes that passes through the points $(2,0)\text{,}$ $(-2,0)\text{,}$ $(0,-1)\text{,}$ and $(0,3)\text{.}$

1. $y=x+1\text{.}$

2. $\ds \frac{x^2}{\frac{16}{3}}+\frac{(y-1)^2}{4}=1\text{.}$

###### 4.

Consider the polar equation $r(3-k\cos \theta )=4\text{.}$

1. For what values of the constant $k$ is this conic section an ellipse?

2. Now assume that $k$ has the value in the middle of the interval found in part (a). Determine the location of the focus and directrix of this conic section.

3. Sketch the graph of this conic section which shows the focus, directrix and vertices.

1. $0\lt k\lt 3\text{.}$

2. The directrix is $\ds x=\frac{8}{3}\text{.}$ The focus is $\frac{8}{9}\text{.}$

3. Solution
1. From $\ds r=\frac{\frac{4}{3}}{1-\frac{k}{3}\cos \theta}$ it follows that this conic section is an ellipse if $0\lt k\lt 3\text{.}$

2. If $\ds k=\frac{3}{2}$ then the eccentricity is given by $\ds e=\frac{k}{3}=\frac{1}{2}\text{.}$ The directrix is $x=d$ where $\ds ed=\frac{4}{3}\text{.}$ Thus the directrix is $\ds x=\frac{8}{3}\text{.}$ Let $c>0$ be such that $(c,0)$ is a focus of the ellipse. Then $\ds \frac{c}{e}(1-e^2)=ed$ and $\ds c=\frac{8}{9}\text{.}$

3. ###### 5.

Assume that the earth is a sphere with radius $s\text{.}$ ($s$ is about 6400 kilometres, but this has nothing to do with the solution of this problem. Express everything, including your answer, in terms of $s\text{.}$) A satellite has an elliptical orbit with the centre of the earth at one focus. The lowest point of the orbit is 5s above the surface of the earth, when the satellite is directly above the North Pole. The highest point of the orbit is 11s above the surface of the earth, when the satellite is directly above the South Pole. What is the height of the satellite above the surface of the earth, when the satellite is directly above the equator? [Suggestion: Choose a coordinate system with the centre of the earth on the $y$-axis and the centre of the satellite's orbit at the origin.]

$8s\text{.}$
Solution

Let the centre of the earth (and a focus of the ellipse) be at the the point $(0,c)\text{,}$ $c>0\text{.}$ Let the equation of the ellpse be $\ds \frac{x^2}{a^2}+ \frac{y^2}{b^2}=1\text{.}$ It is given that the vertices of the ellipse on the $y$-axis are $(0,(c+s)+5s)$ and $(0,(c-s)-11s)\text{.}$ It follows that the length of the major axis on the $y$-axis is $2b=6s+12s=18s\text{.}$ Thus, $b=9s$ and $c=b-6s=3s\text{.}$ From $c^2=b^2-a^2$ we get that $a^2=72s^2\text{.}$ Thus the equation of the ellipse is $\ds \frac{x^2}{72s^2}+ \frac{y^2}{81s^2}=1\text{.}$ The question is to evaluate the value of $|x|$ when $y=c=3s\text{.}$ From $\ds \frac{x^2}{72s^2}+ \frac{9s^2}{81s^2}=1$ it follows that $|x|=8s\text{.}$

###### 6.

Let the focus $F$ of a conic section with eccentricity $e$ be the origin and let the directrix $L$ be the vertical line $x=-p\text{,}$ $p>0\text{.}$ Thus the conic consists of all those points $P=(x,y)$ such that $|PF|=e|PL|\text{.}$ Find the polar equation of the conic section. (With the origin as the pole, and the positive $x$-axis as the polar axis.)

$\ds r=\frac{ep}{1-e\cos \theta }\text{.}$
###### 7.
1. Identify and sketch the graph of the conic section defined by

\begin{equation*} r=\frac{1}{2-4\cos \theta } \end{equation*}

in the $x\text{,}$ $y\text{,}$ plane without converting it to cartesian coordinates. Clearly label all foci and vertices. (Hint: You may use the equation $\ds r=\frac{ep}{1-e\cos \theta }$ without deriving it.)

2. Convert the polar coordinates equation given above to a cartesian coordinates equation of the form $Ax^2+Bx+Cy^2+Dy=E\text{.}$

1. 2. $12x^2-4y^2+8x+1=0\text{.}$

Solution
1. From $\ds r=\frac{\frac{1}{2}}{1-2\cos \theta }$ we see that the eccentricity is $e=2$ and the equation therefore represents a hyperbola. From $\ds ed=\frac{1}{2}$ we conclude that the directrix is $\ds x=-\frac{1}{4}\text{.}$ The vertices occur when $\theta =0$ and $\theta =\pi\text{.}$ Thus the vertices are $\ds \left( -\frac{1}{2},0\right)$ and $\ds \left( -\frac{1}{6},0\right)\text{.}$The $y$-intercepts occur when $\ds \theta =\frac{\pi }{2}$ and $\ds \theta =\frac{3\pi }{2}\text{.}$ Thus $\ds \left( 0, \frac{1}{2}\right)$ and $\ds \left( 0, -\frac{1}{2}\right)\text{.}$ We note that $r\to \infty$ when $\ds \cos \theta \to \frac{1}{2}\text{.}$ Therefore the asymptotes are parallel to the rays $\ds \theta = \frac{\pi }{3}$ and $\ds \theta = \frac{5\pi }{3}\text{.}$ Figure 4.6. $\ds r=\frac{1}{2-4\cos \theta}$ and $\ds x=-\frac{1}{3}$

2. From $\ds \sqrt{x^2+y^2}=\frac{1}{2-\frac{4x}{\sqrt{x^2+y^2}}}$ it follows that the conic section is given by $12x^2-4y^2+8x+1=0\text{.}$

###### 8.
1. Find an equation of the set of all points $(x,y)$ that satisfy this condition: The distance from $(x,y)$ to $(5,0)$ is exactly half the distance from $(x,y)$ to the line $x=-5\text{.}$

2. Simplify your answer from part (a) enough to be able to tell what type of conic section it is.

1. $\ds (x-5)^2+y^2=\frac{(x+5)^2}{4}\text{.}$

2. $\ds 3\left( x-\frac{25}{3}\right) ^2+4y^2=\frac{400}{3}\text{.}$ This is an ellipse.

###### 9.

Derive the equation of the set of all points $P(x,y)$ that are equidistant from the point $A(1,0)$ and the line $x=-5\text{.}$ Provide a diagram with your work. Simplify the equation.

From $(x-1)^2+y^2=(x+5)^2$ we get that $y^2=24+12x\text{.}$ Figure 4.7. $y^2=24+12x$
###### 10.

Let $C$ be the conic section described by the equation

\begin{equation*} x^2-2y^2+4x=0\text{.} \end{equation*}
1. Using the method of completing the square, identify the conic section $C\text{.}$

2. Sketch a graph of $C\text{.}$ Find, foci and asymptotes (if any).

1. $\ds \frac{(x+2)^2}{4}-\frac{y^2}{2}=1\text{.}$ This is a hyperbola.

2. Foci are $(-2-\sqrt{6},0)$ and $(-2+\sqrt{6},0)\text{.}$ The asymptotes are $\ds y=\pm \frac{\sqrt{2}}{2}(x+2)\text{.}$ Figure 4.8. $x^2-2y^2+4x=0$

###### 11.

Given is the polar equation $\ds r=\frac{2}{1-2\cos \theta }\text{.}$

1. Which type of conic section does this polar equation represent: Parabola, ellipse, or hyperbola?

2. Show that the polar equation implies the following equation in Cartesian coordinates:

\begin{equation*} 9\left( x+\frac{4}{3}\right) ^2-3y^2=4\text{.} \end{equation*}
3. Give the foci, vertices, and asymptotes of the conic if it has any.

4. Sketch the conic section based on the information found above. Indicate the features you found in (c) in your sketch.

2. The foci are given by $\ds \left( -\frac{4}{3}\pm \frac{4}{3}, 0\right)\text{,}$ the vertices are given by $\ds \left( -\frac{4}{3}\pm \frac{2}{3}, 0\right)\text{,}$ and the asymptotes are given by $\ds y=\pm \sqrt{3}\left( x+\frac{4}{3}\right)\text{.}$
3. (a) A hyperbola, since the eccentricity is $e=2>1\text{.}$ (b) From $r(1-2\cos \theta)=2$ conclude that $\sqrt{x^2+y^2}=2(x+1)\text{.}$ Square both sides, rearrange the expression, and complete the square.