The extremities of the latera recta of the el | vedclass

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(D) The coordinates of the extremities of the latera recta are given by $(h, k) = (\pm ae, \pm \frac{b^2}{a})$.Since $b^2 = a^2(1 - e^2)$,we have $k =

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Both $(A)$ and $(B)$

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(D) The coordinates of the extremities of the latera recta are given by $(h, k) = (\pm ae, \pm \frac{b^2}{a})$.Since $b^2 = a^2(1 - e^2)$,we have $k = \pm \frac{a^2(1 - e^2)}{a} = \pm a(1 - e^2)$.From $h = \pm ae$,we have $e^2 = \frac{h^2}{a^2}$.Substituting this into the expression for $k$:$k = \pm a(1 - \frac{h^2}{a^2}) = \pm (a - \frac{h^2}{a})$.For the positive sign: $k = a - \frac{h^2}{a} \implies \frac{h^2}{a} = a - k \implies h^2 = a(a - k)$.For the negative sign: $k = -(a - \frac{h^2}{a}) = -a + \frac{h^2}{a} \implies \frac{h^2}{a} = a + k \implies h^2 = a(a + k)$.Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 = a(a - y)$ and $x^2 = a(a + y)$.

The extremities of the latera recta of the ellipses $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ having a given major axis $2a$ lie on:

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