The equation of the ellipse whose foci are (p | vedclass

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(A) The foci are given by $(\pm ae, 0) = (\pm 5, 0)$,so $ae = 5$.The equation of the directrix is $x = \frac{a}{e} = \frac{36}{5}$.Multiplying the two

Vedclass · Accepted Answer

$\frac{x^2}{36} + \frac{y^2}{11} = 1$

Vedclass · Answer

(A) The foci are given by $(\pm ae, 0) = (\pm 5, 0)$,so $ae = 5$.The equation of the directrix is $x = \frac{a}{e} = \frac{36}{5}$.Multiplying the two equations: $(ae) 	imes (\frac{a}{e}) = 5 	imes \frac{36}{5} \implies a^2 = 36 \implies a = 6$.Substituting $a = 6$ into $ae = 5$,we get $6e = 5 \implies e = \frac{5}{6}$.Using the relation $b^2 = a^2(1 - e^2)$,we have $b^2 = 36(1 - \frac{25}{36}) = 36(\frac{11}{36}) = 11$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{36} + \frac{y^2}{11} = 1$.

The equation of the ellipse whose foci are $(\pm 5, 0)$ and one of its directrices is $5x = 36$ is:

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