Let e_1 be the eccentricity of the hyperbola x^2 16 -y^2 9 =1 and… - Vidyadip

MCQ

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

A
$4 \sqrt{5}$
B
$\frac{8 \sqrt{5}}{3}$
✓
$\frac{10 \sqrt{5}}{3}$
D
$3 \sqrt{5}$

✓

Answer

Correct option: C.

$\frac{10 \sqrt{5}}{3}$

c
$ 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 \text { and } b=3 $

$ E: \frac{x^2}{25}+\frac{y^2}{9}=1$

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

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

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