Consider the hyperbola having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and , then is ___________. [2025]

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Solution

Q.
Consider the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having one of its focus at P(–3, 0). If the latus rectum through its other focus subtends a right angle at P and $a^{2} b^{2} = α \sqrt{2} - β, α, β \in N$ , then $α + β$ is ___________. [2025]

Ans.
(1944)

We have, $F_{1} \equiv (- a e, 0) \equiv P (- 3, 0) \Rightarrow a e = 3$

$L_{1} L_{2} = \frac{2 b^{2}}{a}$

In $△ F_{1} F_{2} L_{2}$ ,

$\tan 45 ° = \frac{L_{2} F_{2}}{F_{1} F_{2}} = \frac{b^{2} / a}{2 a e}$

$\Rightarrow 2 a e = \frac{b^{2}}{a}$

$\Rightarrow b^{2} = 6 a$ [ $∵$ ae = 3]

Also, $a^{2} e^{2} = a^{2} + b^{2}$

$\Rightarrow 9 = a^{2} + 6 a$ [ $∵ a e = 3 and b^{2} = 6 a$ ]

$\Rightarrow a^{2} + 6 a - 9 = 0$

$a = - 3 \pm 3 \sqrt{2} = - 3 (1 \pm \sqrt{2})$

Now, $a^{2} b^{2} = a^{2} \cdot 6 a = 6 a^{3}$ [ $∵ b^{2} = 6 a$ ]

$= 6 (135 \sqrt{2} - 189)$

On comparing, we get $α$ = 810 and $β$ = 1134

$∴ α + β = 1944$ .

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