Let , a b and . Let the distane between the foci of E and the foci of H be . If a – A = 2, and the ratio of the eccentricities of E and H is , then the sum of the lengths of their latus rectums is equal to: [2025]

Question

Let $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , a > b and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ . Let the distane between the foci of E and the foci of H be $2 \sqrt{3}$ . If a – A = 2, and the ratio of the eccentricities of E and H is $\frac{1}{3}$ , then the sum of the lengths of their latus rectums is equal to: [2025]

Smart Achievers · Accepted Answer

(4)

We have, $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ , whose foci are ( $\pm$ ae, 0) and $H : \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ , whose foci are ( $\pm$ Ae', 0).

Given, $2 a e = 2 \sqrt{3} and 2 A e' = 2 \sqrt{3}$

$\Rightarrow a e = \sqrt{3} and A e' = \sqrt{3}$

$\Rightarrow \frac{a e}{A e'} = 1 \Rightarrow \frac{e}{e'} = \frac{A}{a}$

$\Rightarrow \frac{1}{3} = \frac{A}{a} \Rightarrow a = 3 A$ $[Given \frac{e}{e'} = \frac{1}{3}]$

Now, a – A = 2 $\Rightarrow$ 2A = 1 [a = 3A]

$\Rightarrow A = 1, a = 3$

$∴ e = \frac{1}{\sqrt{3}} and e' = \sqrt{3}$

Also, $b^{2} = a^{2} (1 - e^{2}) and B^{2} = A^{2} ({(e')}^{2} - 1)$

$\Rightarrow b^{2} = 9 (1 - {(\frac{1}{\sqrt{3}})}^{2}) and B^{2} = {(1)}^{2} ({(\sqrt{3})}^{2} - 1)$

$\Rightarrow b^{2} = 6 and B^{2} = 2$

$∴$ Sum of latus rectum $= \frac{2 b^{2}}{a} + \frac{2 B^{2}}{A} = 8$ .

Solution

1 9

2 7

3 10

4 8

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Solution

Q. Let E : x2a2+y2b2=1, a > b and H : x2A2–y2B2=1. Let the distane between the foci of E and the foci of H be 23. If a – A = 2, and the ratio of the eccentricities of E and H is 13, then the sum of the lengths of their latus rectums is equal to: [2025]

1 9

2 7

3 10

4 8

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