JEE MAIN - Mathematics (2024 - 4th April Evening Shift - No. 1)
Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36 \pi$$ and $$4 \pi$$, respectively, then the length (in units) of latus rectum of $$\mathrm{H}$$ is
$$\frac{28}{3}$$
$$\frac{11}{3}$$
$$\frac{14}{3}$$
$$\frac{10}{3}$$
Explanation
_4th_April_Evening_Shift_en_1_1.png)
$$\begin{aligned} & C_1: x^2+y^2=a^2 \Rightarrow \text { area }=\pi a^2=36 \pi \Rightarrow a=6 \\ & C_2: x^2+(y-a e)^2=(a e-a)^2 \\ & \therefore \quad \pi(a e-a)^2=4 \pi \\ & \Rightarrow 36(e-1)^2=4 \\ & \Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3} \\ & \Rightarrow b^2=28 \\ & \therefore \quad L R=\frac{2 b^2}{a}=\frac{2 \times 28}{6}=\frac{28}{3} \text { units } \end{aligned}$$
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