Jee Main 2024 27 01 2024 Shift 2 - Question11
Question 11
If $H: \frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ and $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b)$. Ellipse passes through the foci of the hyperbola and $e _1 . e _2=1$ (where $e _1, e _2$ are the eccentricities of hyperbola and ellipse, respectively). The length of the chord of ellipse passing through $(0,2)$ is equal to (1) $\frac{5 \sqrt{10}}{3}$ (2) $\frac{10 \sqrt{5}}{3}$ (3) $2 \sqrt{5}$ (4) $2 \sqrt{10}$
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Answer: (2)
Solution:
$e _1^{2}=1+\frac{9}{16} \Rightarrow e _1=\frac{5}{4}$
$\therefore \quad e _2=\frac{4}{5}$
Foci of hyperbola $\equiv( \pm 5,0)$
$2 a=10 \Rightarrow a=5$
$1-\frac{b^{2}}{a^{2}}=\frac{16}{25}$
$\Rightarrow b=3$
$E: \frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
$\frac{x^{2}}{25}+\frac{4}{9}=1 \ldots(y=2)$
$\Rightarrow \frac{x^{2}}{25}=\frac{5}{9}$
$x= \pm \frac{5 \sqrt{5}}{3}$
$\therefore \quad x _1=-\frac{5 \sqrt{5}}{3}, \quad x _2=\frac{5 \sqrt{5}}{3}$
$\therefore \quad I=\frac{10 \sqrt{5}}{3}$
Option (2) is correct.
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