7. If $ heta$ is the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos heta=\frac{5}{13}$, then $a^2=$

Question

7. If $	heta$ is the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos 	heta=\frac{5}{13}$, then $a^2=$

Aicharya · Accepted Answer

Eqn. of hyperbola is $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$ and $\cos 	heta=\frac{5}{13}$ $\Rightarrow \quad 	an 	heta=\frac{12}{5}$ ...(i) Angle between the asymptotes of hyperbola is $	heta=	an ^{-1}\left(\frac{2 	imes 2 a}{a^2-4}ight)$ $\Rightarrow \quad 	an (	heta)=\frac{4 a}{a^2-4}$ ...(ii) From eqns. (i) and (ii) $\begin{aligned} & \frac{12}{5}=\frac{4 a}{a^2-4} \Rightarrow 3 a^2-5 a-12=0 \ & \Rightarrow \quad(a-3)(3 a+4)=0 \ & \Rightarrow \quad a=-\frac{4}{3}, 3 \Rightarrow a^2=\frac{16}{9}, 9 .\end{aligned}$

7. If $\theta$ is the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$…

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