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(B) For the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$,we have $a^2=36$ and $b^2=27$. The foci are $(\pm ae, 0)$.Since $e^2 = 1 - \frac{b^2}{a^2} = 1 -

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$27$

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(B) For the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$,we have $a^2=36$ and $b^2=27$. The foci are $(\pm ae, 0)$.Since $e^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{27}{36} = 1 - \frac{3}{4} = \frac{1}{4}$,we have $e = \frac{1}{2}$.The foci are $(\pm 6 	imes \frac{1}{2}, 0) = (\pm 3, 0)$.$A$ well-known property of an ellipse is that the product of the perpendicular distances from the foci to any tangent is equal to the square of the semi-minor axis,$b^2$.Here,$b^2 = 27$.Thus,the product of the perpendicular distances from the foci $(3,0)$ and $(-3,0)$ to the tangent at any point is $27$.

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at the point $\left(3, \frac{9}{2}\right)$ is:

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