JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Hyperbola

Let the hyperbola

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

and the ellipse

E : 3 x^{2} + 4 y^{2} = 12

be such that the length of latus rectum of H is equal to the length of latus rectum of E. If

e_{H}

and

e_{E}

are the eccentricities of H and E respectively, then the value of

12 (e_{H}^{2} + e_{E}^{2})

is equal to ____.

Answer: 42

Solution

For the hyperbola

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

, we have

b^{2} = 1

, so the latus rectum is

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

.

For the ellipse

E : \frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

, we have

a_{E}^{2} = 4, b_{E}^{2} = 3

, so the latus rectum is

\frac{2 b _{E}^{2}}{a _{E}} = \frac{2 \cdot 3}{2} = 3

.

Equating the latus recta:

\frac{2}{a} = 3 \Rightarrow a = \frac{2}{3}

.

For the hyperbola:

e_{H} = 1 + \frac{b ^{2}}{a ^{2}} = 1 + \frac{1}{4/9} = 1 + \frac{9}{4} = \frac{13}{2}

For the ellipse:

e_{E} = 1 - \frac{b _{E}^{2}}{a _{E}^{2}} = 1 - \frac{3}{4} = \frac{1}{2}

Therefore:

12 (e_{H}^{2} + e_{E}^{2}) = 12 (\frac{13}{4} + \frac{1}{4}) = 12 \cdot \frac{14}{4} = 42

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Integer answer