Let be the hyperbola, whose foci are and eccentricity is . Then the length of its latus rectum is:

2 Explanation: Solution Is question ko solve karne ke liye hum hyperbola ki standard properties ka use karenge: 1. Foci ke beech ki distance ( ): Dono foci aur ke beech ki doori nikaalte hain: Isse humein milta hai: 2. ki value nikalna: Humein eccentricity di gayi hai. mein ki value rakhne par: 3. ki value nikalna: Hyperbola ke liye formula hota hai : 4. Length of Latus Rectum ( ): Latus rectum ka formula hota hai : Correct Option: (2)

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JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $H$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$ . Then the length of its latus rectum is:

Choose the correct answer:

Explanation

Solution

Is question ko solve karne ke liye hum hyperbola ki standard properties ka use karenge:

1. Foci ke beech ki distance ( $2ae$ ):

Dono foci $(1 + \sqrt{2}, 0)$ aur $(1 - \sqrt{2}, 0)$ ke beech ki doori nikaalte hain:

$2ae = (1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2}$

Isse humein milta hai:

$ae = \sqrt{2}$

2. $a$ ki value nikalna:

Humein eccentricity $e = \sqrt{2}$ di gayi hai. $ae = \sqrt{2}$ mein $e$ ki value rakhne par:

$a(\sqrt{2}) = \sqrt{2} \implies a = 1$

3. $b^2$ ki value nikalna:

Hyperbola ke liye formula hota hai $b^2 = a^2(e^2 - 1)$ :

$b^2 = 1^2((\sqrt{2})^2 - 1)$

$b^2 = 1(2 - 1) = 1$

4. Length of Latus Rectum ( $L$ ):

Latus rectum ka formula hota hai $\frac{2b^2}{a}$ :

$L = \frac{2(1)}{1} = 2$

Correct Option: (2)

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