JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Hyperbola

Let the foci of a hyperbola H coincide with the foci of the ellipse

E : \frac{( x - 1 ) ^{2}}{100} + \frac{( y - 1 ) ^{2}}{75} = 1

and the eccentricity of the hyperbola H be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of H is

α

and the length of its conjugate axis is

β

, then

3 α^{2} + 2 β^{2}

is equal to :

Answer: (B)

Solution

Ellipse

E : \frac{( x - 1 ) ^{2}}{100} + \frac{( y - 1 ) ^{2}}{75} = 1

a^{2} = 100, b^{2} = 75

, so

e_{1} = 1 - \frac{75}{100} = \frac{1}{2}, c = 10 \times \frac{1}{2} = 5

Foci of

E

(1 + 5, 1) = (6, 1)

and

(1 - 5, 1) = (- 4, 1)

.

Hyperbola

H

shares these foci, so its centre is

(1, 1)

and

a_{H} e_{2} = 5

, with

e_{2} = 1/ e_{1} = 2 ⟹ a_{H} = \frac{5}{2}

. Transverse axis:

α = 2 a_{H} = 5

.

For a hyperbola,

b_{H}^{2} = a_{H}^{2} (e_{2}^{2} - 1) = \frac{25}{4} (4 - 1) = \frac{75}{4}

, so conjugate axis

β = 2 b_{H} = 53

3 α^{2} + 2 β^{2} = 3 (5)^{2} + 2 (53)^{2} = 3 (25) + 2 (75) = 75 + 150 = 225

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option