If theta is the angle between the asymptotes | vedclass

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(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$. Given $\cos 	heta = \frac{5}{13}$,we have $	an 	heta = \frac{12}{5}$ (a

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$\frac{16}{9}$ or $9$

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(B) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{(y-2)^2}{4}=1$. Given $\cos 	heta = \frac{5}{13}$,we have $	an 	heta = \frac{12}{5}$ (assuming $	heta$ is acute). The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ are $y-k = \pm \frac{b}{a}x$. The slopes are $m_1 = \frac{b}{a}$ and $m_2 = -\frac{b}{a}$. The angle $	heta$ between the asymptotes is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{2b/a}{1 - b^2/a^2} ight| = \left| \frac{2ab}{a^2 - b^2} ight|$. Here $b^2 = 4$,so $b = 2$. Thus,$	an 	heta = \left| \frac{4a}{a^2 - 4} ight|$. Equating the two expressions for $	an 	heta$: $\frac{12}{5} = \frac{4a}{a^2 - 4}$ or $\frac{12}{5} = -\frac{4a}{a^2 - 4}$. Case $1$: $3(a^2 - 4) = 5a$ $\Rightarrow 3a^2 - 5a - 12 = 0$ $\Rightarrow (3a + 4)(a - 3) = 0$. Since $a^2 > 0$,$a = 3 \Rightarrow a^2 = 9$. Case $2$: $3(a^2 - 4) = -5a$ $\Rightarrow 3a^2 + 5a - 12 = 0$ $\Rightarrow (3a - 4)(a + 3) = 0$. Since $a^2 > 0$,$a = 4/3 \Rightarrow a^2 = 16/9$. Thus,$a^2 = \frac{16}{9}$ or $9$.

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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