A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Section 6.16 Conic Sections

1. See Figure 6.16.1. Focci: $(0,-\sqrt{7})\text{,}$ $(0,\sqrt{7})\text{.}$

   

2. (a) $\ds e=\frac{1}{2}\text{.}$ (b) Use the fact that, for $P=(x,y)\text{,}$ $\ds |PF|^2=x^2+(y-1)^2$ and $\ds |Pl|=\frac{1}{2}|y-4|\text{.}$ (c) 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. (a) $y=x+1\text{.}$ (b) $\ds \frac{x^2}{\frac{16}{3}}+\frac{(y-1)^2}{4}=1\text{.}$

4. (a) From $\ds r=\frac{\frac{4}{3}}{1-\frac{k}{3}\cos \theta}$ it follows that this conic section is an ellipse if $0\lt k\lt 3\text{.}$ (b) 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 $\ds c=\frac{8}{9}\text{.}$ (c) See Figure 6.16.2.

   

5. (a) 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 $|x|=8s\text{.}$ (b) $\ds r=\frac{ep}{1-e\cos \theta }\text{.}$

6. (a) 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 $\ds \theta = \frac{5\pi }{3}\text{.}$ See Figure 6.16.3.

   

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

7. (a) $\ds (x-5)^2+y^2=\frac{(x+5)^2}{4}\text{.}$ (b) $\ds 3\left( x-\frac{25}{3}\right) ^2+4y^2=\frac{400}{3}\text{.}$ This is an ellipse.

8. From $(x-1)^2+y^2=(x+5)^2$ we get that $y^2=24+12x\text{.}$ See Figure 6.16.4.

   

9. (a) $\ds \frac{(x+2)^2}{4}-\frac{y^2}{2}=1\text{.}$ This is a hyperbola. Foci are $(-2-\sqrt{6},0)$ and $(-2+\sqrt{6},0)\text{.}$ The asymptotes are $\ds y=\pm \frac{\sqrt{2}}{2}(x+2)\text{.}$ (b) See Figure 6.16.5.

   

10. (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. (c) From $\ds \frac{\left( x+\frac{4}{3}\right) ^2}{\left( \frac{2}{3}\right) ^2}-\frac{y^2}{\left( \frac{2}{\sqrt{3}}\right) ^2}=1$ it follows that 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 $\ds y=\pm \sqrt{3}\left( x+\frac{4}{3}\right)\text{.}$ (d) See Figure 6.16.6.