Extremities of the latera recta of the ellips | vedclass

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Question

$h = \pm ae ; k = \pm \frac{{{b^2}}}{a}\,$
$k = \pm a(1 -e^2) = \pm  a \left( {1 - \frac{{{h^2}}}{{{a^2}}}} \right)\,$ $=$ $ \pm$  $\left(

Vedclass · Accepted Answer

both $(A)$ and $(B)$

Vedclass · Answer

$h = \pm ae ; k = \pm \frac{{{b^2}}}{a}\,$
$k = \pm a(1 -e^2) = \pm  a \left( {1 - \frac{{{h^2}}}{{{a^2}}}} \right)\,$ $=$ $ \pm$  $\left( {a - \frac{{{h^2}}}{a}} \right)\,$ 
$+ ve\ sign , k =$ $a - \frac{{{h^2}}}{a}\,\,$ $\Rightarrow$ $\frac{{{h^2}}}{a}\,\, = \,a - k$ $\Rightarrow$ $h^2 = a ( a - k)$ $\Rightarrow$ $(A)$ 
$- ve\ sign , k = - a + \frac{{{h^2}}}{a}\,\,$  $\Rightarrow$ $h^2 = a (a + k)$

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 $2a$ lies on

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