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

These answers correspond to the problems in Section 4.4.

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.