If the latus rectum through one of the foci o | vedclass

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(D) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$,so $a = 3$. The foci are $(\pm ae, 0)$ and vertices are $(\pm a, 0)$.L

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$27$

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(D) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$,so $a = 3$. The foci are $(\pm ae, 0)$ and vertices are $(\pm a, 0)$.Let the focus be $S(ae, 0)$ and the farther vertex be $A'(-a, 0)$.The latus rectum is the line $x = ae$. The endpoints of the latus rectum are $L(ae, \frac{b^2}{a})$ and $L'(ae, -\frac{b^2}{a})$.The vector $\vec{A'L} = (ae - (-a), \frac{b^2}{a} - 0) = (a(e+1), \frac{b^2}{a})$.The vector $\vec{A'L'} = (ae - (-a), -\frac{b^2}{a} - 0) = (a(e+1), -\frac{b^2}{a})$.Since the angle $\angle L A' L'$ is $90^\circ$,the dot product $\vec{A'L} \cdot \vec{A'L'} = 0$.$a^2(e+1)^2 - \frac{b^4}{a^2} = 0 \implies a^4(e+1)^2 = b^4$.Since $b^2 = a^2(e^2-1) = a^2(e-1)(e+1)$,we have $b^4 = a^4(e-1)^2(e+1)^2$.Equating the two expressions for $b^4$: $a^4(e+1)^2 = a^4(e-1)^2(e+1)^2$.Dividing by $a^4(e+1)^2$,we get $1 = (e-1)^2$,so $e-1 = 1$,which means $e = 2$.Then $b^2 = a^2(e^2-1) = 9(2^2-1) = 9(3) = 27$.

If the latus rectum through one of the foci of a hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtends a right angle at the farther vertex of the hyperbola,then $b^2=$

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