Let $\mathrm{e}_{1}$ be the eccentricity of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ and $e_{2}$ be the eccentricity of the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}>\mathrm{b}$, 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 :

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Ellipse

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 :

Answer: (C)

For the hyperbola

\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1

a^{2} = 16, b^{2} = 9

, so

e_{1} = 1 + \frac{9}{16} = \frac{5}{4},

with foci at

(\pm 5, 0)

.

Given

e_{1} e_{2} = 1

e_{2} = \frac{4}{5}

.

The ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

(

a > b

) passes through the hyperbola's foci

(\pm 5, 0)

, so its semi-major axis is

a = 5

. From

e_{2} = 1 - \frac{b ^{2}}{a ^{2}} = \frac{4}{5}

1 - \frac{b ^{2}}{25} = \frac{16}{25} \Rightarrow b^{2} = 9.

The ellipse is

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

. The chord through

(0, 2)

parallel to the x-axis has

y = 2

\frac{x ^{2}}{25} + \frac{4}{9} = 1 \Rightarrow x^{2} = 25 (1 - \frac{4}{9}) = \frac{125}{9} \Rightarrow x = \pm \frac{5 5}{3} .

Chord length

= 2 \times \frac{5 5}{3} = \frac{10 5}{3}

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Single correct option