Let $H_{1}: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $H_{2}:-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$ be two hyperbolas having length of latus rectums $15 \sqrt{2}$ and $12 \sqrt{5}$ respectively. Let their eccentricities be $\mathrm{e}_{1}=\sqrt{\frac{5}{2}}$ and $\mathrm{e}_{2}$ respectively. If the product of the lengths of their transverse axes is $100 \sqrt{10}$, then $25 \mathrm{e}_{2}^{2}$ is equal to ____

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Hyperbola

Let

H_{1} : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

and

H_{2} : - \frac{x ^{2}}{A ^{2}} + \frac{y ^{2}}{B ^{2}} = 1

be two hyperbolas having length of latus rectums

152

and

125

respectively. Let their eccentricities be

e_{1} = \frac{5}{2}

and

e_{2}

respectively. If the product of the lengths of their transverse axes is

10010

, then

25 e_{2}^{2}

is equal to ____

Answer: 55

For

H_{1} : \frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

: eccentricity gives

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

, so

\frac{b ^{2}}{a ^{2}} = \frac{3}{2}

, i.e.

b^{2} = \frac{3}{2} a^{2}

.

The latus rectum

\frac{2 b ^{2}}{a} = 152

gives

\frac{2 \cdot \frac{3}{2} a ^{2}}{a} = 3 a = 152 \Rightarrow a = 52, b^{2} = \frac{3}{2} (50) = 75.

So the transverse axis of

H_{1}

has length

2 a = 102

.

For

H_{2} : \frac{y ^{2}}{B ^{2}} - \frac{x ^{2}}{A ^{2}} = 1

(the given form with

A^{2}, B^{2}

playing the roles of the transverse/conjugate semi-axes along

y, x

), the transverse axis length is

2 B

.

Given the product of transverse axes is

10010

(2 a) (2 B) = 10010 \Rightarrow 102 \cdot 2 B = 10010 \Rightarrow B = \frac{100 10}{20 2} = 55, B^{2} = 125.

The latus rectum of

H_{2}

\frac{2 A ^{2}}{B} = 125

2 A^{2} = 125 \cdot 55 = 300 \Rightarrow A^{2} = 150.

For

H_{2}

e_{2}^{2} = 1 + \frac{A ^{2}}{B ^{2}} = 1 + \frac{150}{125} = \frac{11}{5}

.

Therefore

25 e_{2}^{2} = 25 \cdot \frac{11}{5} = 55.

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Integer answer