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7. If $\theta$ is the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos \theta=\frac{5}{13}$, then $a^2=$
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The correct answer is:
$\frac{16}{9}$ or 9
Eqn. of hyperbola is
$\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos \theta=\frac{5}{13}$
$\Rightarrow \quad \tan \theta=\frac{12}{5}$ ...(i)
Angle between the asymptotes of hyperbola is
$\theta=\tan ^{-1}\left(\frac{2 \times 2 a}{a^2-4}\right)$
$\Rightarrow \quad \tan (\theta)=\frac{4 a}{a^2-4}$ ...(ii)
From eqns. (i) and (ii)
$\begin{aligned} & \frac{12}{5}=\frac{4 a}{a^2-4} \Rightarrow 3 a^2-5 a-12=0 \\ & \Rightarrow \quad(a-3)(3 a+4)=0 \\ & \Rightarrow \quad a=-\frac{4}{3}, 3 \Rightarrow a^2=\frac{16}{9}, 9 .\end{aligned}$
$\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos \theta=\frac{5}{13}$
$\Rightarrow \quad \tan \theta=\frac{12}{5}$ ...(i)
Angle between the asymptotes of hyperbola is
$\theta=\tan ^{-1}\left(\frac{2 \times 2 a}{a^2-4}\right)$
$\Rightarrow \quad \tan (\theta)=\frac{4 a}{a^2-4}$ ...(ii)
From eqns. (i) and (ii)
$\begin{aligned} & \frac{12}{5}=\frac{4 a}{a^2-4} \Rightarrow 3 a^2-5 a-12=0 \\ & \Rightarrow \quad(a-3)(3 a+4)=0 \\ & \Rightarrow \quad a=-\frac{4}{3}, 3 \Rightarrow a^2=\frac{16}{9}, 9 .\end{aligned}$
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