Let be the eccentricity of the hyperbola and be the eccentricity of the ellipse , a > b, which passes through the foci of the hyperbola. If , then the length of the chord of the ellipse parallel to the x-axis and passing through (0, 2) is [202

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Solution

Q.
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 x-axis and passing through (0, 2) is [2024]

1 $4 \sqrt{5}$

2 $\frac{10 \sqrt{5}}{3}$

3 $\frac{8 \sqrt{5}}{3}$

4 $3 \sqrt{5}$

Ans.
(2)

Given hyperbola is $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$

$e_{1}$ = eccentricity of hyperbola = $\frac{\sqrt{16 + 9}}{4} = \frac{5}{4}$

Foci of hyperbola = ( $\pm$ 5, 0)

Given $e_{1} e_{2} = 1$

$\Rightarrow e_{2} = \frac{4}{5}$ = eccentricity of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow$ $\frac{\sqrt{a^{2} - b^{2}}}{a} = \frac{4}{5}$ ... (i)

$∴$ The ellipse also passes through the foci of the hyperbola.

$\Rightarrow a^{2} = 25$

From (i), $b^{2} = 9$

Thus, the equation of ellipse is $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$ ... (ii)

Now, equation of chord of ellipse which passes through (0, 2) and parallel to x-axis is y = 2.

From (ii),

$\frac{x^{2}}{25} = \frac{5}{9} \Rightarrow x^{2} = \frac{25 \times 5}{9} \Rightarrow x = \pm \frac{5 \sqrt{5}}{3}$

Thus, the end point of chord are $(\frac{- 5 \sqrt{5}}{3}, 2) and (\frac{5 \sqrt{5}}{3}, 2)$ .

$∴$ Length of chord = $\sqrt{{(\frac{5 \sqrt{5}}{3} + \frac{5 \sqrt{5}}{3})}^{2} + {(2 - 2)}^{2}} = \frac{10 \sqrt{5}}{3}$ .

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