Let , and a = fog (10), b = gof (3). If e and denote the eccentricity and the length of the latus rectum of the ellipse , then + is equal to [2024]

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Solution

Q.
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 [2024]

1 6

2 12

3 16

4 8

Ans.
(4)

We have, $g (x) = \frac{x}{x - 9} so, g (10) = 10$

Also $f (x) = x^{2} + 9 so, f (10) = 10^{2} + 9 = 109$

$a = f o g (10) = f (g (10)) = f (10) = 109$

Also, $f (3) = 3^{2} + 9 = 18$

Now, $b = g o f (3) = g (f (3)) = g (18) = \frac{18}{18 - 9} = 2$

So ellipse $\frac{x^{2}}{a} + \frac{y^{2}}{b} = 1$ becomes $\frac{x^{2}}{109} + \frac{y^{2}}{2} = 1$

$∴ e = \sqrt{1 - \frac{2}{109}} = \sqrt{\frac{107}{109}}$

and $l$ $= \frac{2 b^{2}}{a} = \frac{2 \times 2}{\sqrt{109}} = \frac{4}{\sqrt{109}}$

Hence, $8 e^{2}$ + $l^{2}$ $= \frac{8 \times 107}{109} + \frac{16}{109} = 8$ .

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