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QJEEM24Macb05: JEE Main 2024 Mathematics - Class 11 JEEM24Macb05 Solution: Let be the focus of the hyperbola , on the positive -axis. Let…

Let S\mathrm{S} be the focus of the hyperbola x23−y25=1\frac{x^2}{3}-\frac{y^2}{5}=1 , on the positive xx -axis. Let C\mathrm{C} be the circle with its centre at A(6,5)\mathrm{A}(\sqrt{6}, \sqrt{5}) and passing through the point S\mathrm{S} . If O\mathrm{O} is the origin and SAB\mathrm{SAB} is a diameter of C\mathrm{C} , then the square of the area of the triangle OSB is equal to __________.

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amp;x23−y25=1amp;5=3(e2−1)⇒e=83amp;S≡(22,0)\begin{aligned}& \frac{x^2}{3}-\frac{y^2}{5}=1 \\& 5=3\left(e^2-1\right) \Rightarrow e=\sqrt{\frac{8}{3}} \\& S \equiv(2 \sqrt{2}, 0)\end{aligned}

JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Hyperbola Question 5 English Explanation

AA is mid-point of BSB S

⇒B(26−22,25)\Rightarrow \quad B(2 \sqrt{6}-2 \sqrt{2}, 2 \sqrt{5})

Δ(OSB)=|12|001220126−22251||=210<./p>

(Δ(OSB))2=40(\Delta(O S B))^2=40

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