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JEE Main 2024 (Online) 8th April Evening Shift
Numerical
+4
-1

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

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2
JEE Main 2024 (Online) 6th April Evening Shift
Numerical
+4
-1

The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $$x= \pm \frac{4}{\sqrt{3}}$$, respectively. Let the line $$y-\sqrt{3} x+\sqrt{3}=0$$ touch this hyperbola at $$\left(x_0, y_0\right)$$. If $$\mathrm{m}$$ is the product of the focal distances of the point $$\left(x_0, y_0\right)$$, then $$4 \mathrm{e}^2+\mathrm{m}$$ is equal to _________.

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3
JEE Main 2024 (Online) 31st January Morning Shift
Numerical
+4
-1
Change Language

Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$$ and $$\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1$$ equals

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4
JEE Main 2024 (Online) 30th January Morning Shift
Numerical
+4
-1
Change Language

Let the latus rectum of the hyperbola $$\frac{x^2}{9}-\frac{y^2}{b^2}=1$$ subtend an angle of $$\frac{\pi}{3}$$ at the centre of the hyperbola. If $$\mathrm{b}^2$$ is equal to $$\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$$, where $$l$$ and $$\mathrm{m}$$ are co-prime numbers, then $$\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2$$ is equal to ________.

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