Hyperbola Question 3
Question 3 - 2024 (27 Jan Shift 2)
Let $e_{1}$ be the eccentricity of the hyperbola $\frac{\mathrm{x}^{2}}{16}-\frac{\mathrm{y}^{2}}{9}=1$ and $\mathrm{e}{2}$ be the eccentricity of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$, 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 $\mathrm{x}$-axis and passing through $(0,2)$ is :
(1) $4 \sqrt{5}$
(2) $\frac{8 \sqrt{5}}{3}$
(3) $\frac{10 \sqrt{5}}{3}$
(4) $3 \sqrt{5}$
Show Answer
Answer (3)
Solution
$H: \frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \quad e_{1}=\frac{5}{4}$
$\therefore e_{1} e_{2}=1 \Rightarrow e_{2}=\frac{4}{5}$
Also, ellipse is passing through $( \pm 5,0)$
$\therefore a=5$ and $b=3$
$E: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
End point of chord are $\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$
$\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}$
