Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• (a) For each $p>0,$ the equation $px^{2} + (p+2)y^{2} = p^{2} + 2p$ represents an ellipse. Find (in terms of $p$) the eccentricity and the coordinates of the foci.
  (b) Find a Cartesian equation for the hyperbola which has the same foci as the ellipse of part (a) and which has eccentricity $\sqrt{3}.$
• (a) Rearrange the terms of the equation $px^{2} + (p+2)y^{2} = p^{2} + 2p$ to find the standard form of an ellipse. What are $a^{2}$ and $b^{2}$ in terms of $p$?
  (b) Recall the standard form of a hyperbola symmetric about the origin$$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}$$where $b^{2} = a^{2}(e^{2} - 1).$ If the foci are the same as in part (a), but the eccentricity is $\sqrt{3},$ what are $a^{2}$ and $b^{2}\ $?
• (a) Rearranging the terms of the equation $px^{2} + (p+2)y^{2} = p^{2} + 2p,$ we get the standard form $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$$where $a^{2} = p+2,$ $b^{2} = p.$ Substituting $b^{2} = a^{2}(1 - e^{2})$ we find that $(1 - e^{2}) = \frac{p}{p+2}$ or in other words, $$e = \sqrt{\frac{2}{p+2}}$$Since we know that the ellipse is symmetric about the origin with vertical directrices, we know that its foci are at the points $(ae, 0),$ $(-ae, 0),$ or $(\sqrt{2}, 0)$ and $(-\sqrt{2}, 0) \quad \blacksquare$ \begin{align*} \end{align*} (b) The foci of a hyperbola symmetric about the origin with vertical directrices are at the points $(ae, 0)$ and $(-ae, 0).$ From (a) we know that these points are $(\sqrt{2}, 0)$ and $(-\sqrt{2}, 0).$ But if $e = \sqrt{3},$ then $\pm ae = \pm \sqrt{3}a = \pm \sqrt{2},$ or in other words, $$a^{2} = \frac{2}{3}$$giving us \begin{align*}\\b^{2} &= a^{2}(e^{2} - 1)\\&=\frac{4}{3}\end{align*}Substituting this into the Cartesian equation in standard form, we get \begin{align*}\frac{3x^{2}}{2} - \frac{3y^{2}}{4} = 1\end{align*}or, $\quad6x^{2} - 3y^{2} = 4. \quad \blacksquare$