JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Hyperbola

For the hyperbola H :

x^{2} - y^{2} = 1

and the ellipse

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b > 0

, let the
(1) eccentricity of

E

be reciprocal of the eccentricity of H, and
(2) the line

y = \frac{5}{2} x + K

be a common tangent of E and H.

Then

4 (a^{2} + b^{2})

is equal to

Answer: 3

Solution

For hyperbola

H : x^{2} - y^{2} = 1

, eccentricity

e_{H} = 2

.

For ellipse

E

, eccentricity

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

.

Condition (1):

e_{E} = \frac{1}{e _{H}} = \frac{1}{2}

.

This gives

1 - \frac{b ^{2}}{a ^{2}} = \frac{1}{2} \Rightarrow a^{2} = 2 b^{2}

.

For condition (2), the line

y = \frac{5}{2} x + K

is tangent to the ellipse when

K^{2} = a^{2} m^{2} + b^{2}

where

m = \frac{5}{2}

.

Substituting:

K^{2} = 2 b^{2} \cdot \frac{5}{2} + b^{2} = 6 b^{2}

.

For the hyperbola, tangent condition gives

K^{2} = 1 \cdot m^{2} - 1 = \frac{5}{2} - 1 = \frac{3}{2}

.

Therefore,

6 b^{2} = \frac{3}{2} \Rightarrow b^{2} = \frac{1}{4}

and

a^{2} = \frac{1}{2}

.

Thus,

4 (a^{2} + b^{2}) = 4 (\frac{1}{2} + \frac{1}{4}) = 3

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Integer answer