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JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Ellipse

Let

f (x) = x^{2} + 9, g (x) = \frac{x}{x - 9}

and

a = fog (10), b = gof (3)

. If e and l denote the eccentricity and the length of the latus rectum of the ellipse

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

, then

8 e^{2} + l^{2}

is equal to.

(A)16
(B)8✓ Correct
(C)6
(D)12

Answer: (B)

Solution

g (10) = 10 \Rightarrow a = f (10) = 109

;

f (3) = 18 \Rightarrow b = g (18) = 2

. For

\frac{x ^{2}}{109} + \frac{y ^{2}}{2} = 1

(major axis along

x

):

e^{2} = 1 - \frac{2}{109} = \frac{107}{109}

and

l = \frac{2 ( 2 )}{109}

, so

l^{2} = \frac{16}{109}

. Then

8 e^{2} + l^{2} = \frac{856 + 16}{109} = 8

— option (2).

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Single correct option

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