Let H: frac{x^2}{a^2} - frac{y^2}{b^2} = 1 be | vedclass

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(A) Given the distance between foci is $2ae = 6$,so $ae = 3$. The distance between directrices is $\frac{2a}{e} = \frac{8}{3}$,so $\frac{a}{e} = \frac

Vedclass · Accepted Answer

$12$

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(A) Given the distance between foci is $2ae = 6$,so $ae = 3$. The distance between directrices is $\frac{2a}{e} = \frac{8}{3}$,so $\frac{a}{e} = \frac{4}{3}$. Multiplying these,we get $a^2 = 3 	imes \frac{4}{3} = 4$,so $a = 2$. Then $e^2 = \frac{ae}{a/e} = \frac{3}{4/3} = \frac{9}{4}$,so $e = \frac{3}{2}$. Using $b^2 = a^2(e^2 - 1)$,we get $b^2 = 4(\frac{9}{4} - 1) = 4(\frac{5}{4}) = 5$. The hyperbola equation is $\frac{x^2}{4} - \frac{y^2}{5} = 1$. For the line $x = k$,the points of intersection $A$ and $B$ are $(k, y_1)$ and $(k, -y_1)$,where $\frac{k^2}{4} - \frac{y_1^2}{5} = 1$,so $y_1^2 = 5(\frac{k^2}{4} - 1)$. The area of triangle $AOB$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2y_1) 	imes k = k y_1 = 4\sqrt{15}$. Squaring both sides,$k^2 y_1^2 = 16 	imes 15 = 240$. Substituting $y_1^2$,$k^2 	imes 5(\frac{k^2}{4} - 1) = 240 \implies k^2(\frac{k^2 - 4}{4}) = 48 \implies k^4 - 4k^2 - 192 = 0$. Let $u = k^2$,then $u^2 - 4u - 192 = 0 \implies (u - 16)(u + 12) = 0$. Since $k^2 > 0$,$k^2 = 16$,so $k = 4$. Note: The question asks for $a^2$,which is $4$. However,given the options,there might be a typo in the question statement regarding the line $x=a$. If the line was $x=k$ and the result is $a^2=12$,it implies a different set of parameters. Based on the standard calculation,$a^2=4$.

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