JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 26)
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 ________.
Answer
182
Explanation
LR subtends $$60^{\circ}$$ at centre
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$$\begin{aligned} & \Rightarrow \tan 30^{\circ}=\frac{\mathrm{b}^2 / \mathrm{a}}{\mathrm{ae}}=\frac{\mathrm{b}^2}{\mathrm{a}^2 \mathrm{e}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^2}{9} \end{aligned}$$
Also, $$\mathrm{e}^2=1+\frac{\mathrm{b}^2}{9} \Rightarrow 1+\frac{\mathrm{b}^2}{9}=\frac{3 \mathrm{~b}^4}{81}$$
$$\begin{aligned} & \Rightarrow \mathrm{b}^4=3 \mathrm{~b}^2+27 \\ & \Rightarrow \mathrm{b}^4-3 \mathrm{~b}^2-27=0 \\ & \Rightarrow \mathrm{b}^2=\frac{3}{2}(1+\sqrt{13}) \\ & \Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13 \\ & \Rightarrow \ell^2+\mathrm{m}^2+\mathrm{n}^2=182 \end{aligned}$$
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