The equation of the ellipse whose latus rectu | vedclass

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(C) The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.Given the eccentricity $e = \frac{1}{\sqrt{

Vedclass · Accepted Answer

$\frac{x^2}{64} + \frac{y^2}{32} = 1$

Vedclass · Answer

(C) The length of the latus rectum of an ellipse is given by $\frac{2b^2}{a} = 8$,which implies $b^2 = 4a$.Given the eccentricity $e = \frac{1}{\sqrt{2}}$,we use the relation $b^2 = a^2(1 - e^2)$.Substituting $e^2 = \frac{1}{2}$,we get $b^2 = a^2(1 - \frac{1}{2}) = \frac{a^2}{2}$,so $a^2 = 2b^2$.Substituting $b^2 = 4a$ into $a^2 = 2b^2$,we get $a^2 = 2(4a) = 8a$.Since $a > 0$,$a = 8$. Then $a^2 = 64$.Now,$b^2 = 4a = 4(8) = 32$.The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{64} + \frac{y^2}{32} = 1$.

The equation of the ellipse whose latus rectum is $8$ and whose eccentricity is $\frac{1}{\sqrt{2}}$,referred to the principal axes of coordinates,is

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