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Question

(D) The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y = \pm \frac{b}{a}x$. Let $2	heta$ be the angle between the asymptotes.

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(D) The asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y = \pm \frac{b}{a}x$. Let $2	heta$ be the angle between the asymptotes. Then $	an 	heta = \frac{b}{a}$. Given that the angle between the asymptotes is $2 	an^{-1}\left(\frac{2}{3}ight)$,we have $	an 	heta = \frac{2}{3}$,so $\frac{b}{a} = \frac{2}{3}$,which implies $b = \frac{2}{3}a$. Substitute $b^2 = \frac{4}{9}a^2$ into the given equation $a^2 - b^2 = 45$: $a^2 - \frac{4}{9}a^2 = 45$ $\frac{5}{9}a^2 = 45$ $a^2 = 45 	imes \frac{9}{5} = 81$,so $a = 9$. Then $b^2 = \frac{4}{9}(81) = 36$,so $b = 6$. Therefore,$ab = 9 	imes 6 = 54$.

If the angle between the asymptotes of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan^{-1}\left(\frac{b}{a}\right) = 2 \tan^{-1}\left(\frac{2}{3}\right)$ and $a^2-b^2=45$,then $ab=$

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