Let the hyperbola H:x^2a^2-y^2=1 and the ellipse E:3x^2+4y^2=12 be such that the length of…

Question

Let the hyperbola $H:\dfrac{x^2}{a^2}-y^2=1$ and the ellipse $E:3x^2+4y^2=12$ be such that the length of the latus rectum of $H$ equals the length of the latus rectum of $E$. If $e_H$ and $e_E$ are the eccentricities of $H$ and $E$ respectively, then the value of $12(e_H^2+e_E^2)$ is equal to:

Answer: 42

Accepted Answer

$E:\dfrac{x^2}4+\dfrac{y^2}3=1$, latus rectum $=\dfrac{2\cdot3}2=3$. Setting $H$'s latus rectum $\dfrac{2(1)}a=3\Rightarrow a=\dfrac23,\ a^2=\dfrac49$.

$$e_H^2=1+\frac1{a^2}=1+\frac94=\frac{13}4,\qquad e_E^2=1-\frac34=\frac14.$$

$$12(e_H^2+e_E^2)=12\left(\frac{13}4+\frac14\right)=12\cdot\frac{14}4=42.$$

Solutions · 1