JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 8)
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 :
$$\frac{8 \sqrt{5}}{3}$$
$$3 \sqrt{5}$$
$$4 \sqrt{5}$$
$$\frac{10 \sqrt{5}}{3}$$
Explanation
$$\begin{aligned} & H: \frac{x^2}{16}-\frac{y^2}{9}=1 \qquad e_1=\frac{5}{4} \\ & \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5} \end{aligned}$$
Also, ellipse is passing through $$( \pm 5,0)$$
$$\begin{aligned} & \therefore a=5 \text { and } b=3 \\ & E: \frac{x^2}{25}+\frac{y^2}{9}=1 \end{aligned}$$
_27th_January_Evening_Shift_en_8_1.png)
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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