Let S be the focus of the hyperbola frac x 2 3 -frac y 2 5 =1 , on the positive x -axis. Let C be the circle with its centre at A(√(6), √(5)) and passing through the point S . if O is the origin and SAB is a diameter of C then the square of the area of the triangle OSB is equal to :

Mathematics · Hyperbola

JEE Main 2024 — 8 April, Shift 2 — Question 21

Let $S$ be the focus of the hyperbola $\frac{x^{2}}{3}-\frac{y^{2}}{5}=1$ , on the positive x -axis. Let C be the circle with its centre at $A(\sqrt{6}, \sqrt{5})$ and passing through the point S . if O is the origin and SAB is a diameter of $C$ then the square of the area of the triangle OSB is equal to :

Answer: 40

Numerical answer — enter this value.

Step-by-step solution

Area $=\frac{1}{2}(\mathrm{OS})$

$\mathrm{h}=\frac{1}{2} 2\sqrt{8} \sqrt{5}=\sqrt{40}$

$h^2=40$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 2
Subject: Mathematics
Chapter: Hyperbola
Topic: Introduction to Hyperbola