Let e_1 be the eccentricity of the hyperbola [(x^2)/(16)-(y^2)/(9)=1] and e_2 be the eccen…

Question

Let $e_1$ be the eccentricity of the hyperbola $$\frac{x^2}{16}-\frac{y^2}{9}=1$$ and $e_2$ be the eccentricity of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a} > \mathrm{b}$$, which passes through the foci of the hyperbola. If $$\mathrm{e}_1 \mathrm{e}_2=1$$, then the length of the chord of the ellipse parallel to the $x$-axis and passing through $(0,2)$ is :

Accepted Answer

$$\begin{aligned}
& H: \frac{x^2}{16}-\frac{y^2}{9}=1 \qquad e_1=\frac{5}{4} \
& \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5}
\end{aligned}$$

Also, ellipse is passing through $$( \pm 5,0)$$

$$\begin{aligned}
& \therefore a=5 \text { and } b=3 \
& E: \frac{x^2}{25}+\frac{y^2}{9}=1
\end{aligned}$$

End point of chord are $$\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$$

$$\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}$$

Solutions · 1