Let the length of the latus rectum of an elli | vedclass

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Question

Consider $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Given that $2b = 2ae$

$ \Rightarrow b = ae$ and $\frac{{2{b^2}}}

Vedclass · Accepted Answer

$\left( {4,\sqrt 3 ,2\sqrt 2 } \right)$

Vedclass · Answer

Consider $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

Given that $2b = 2ae$

$ \Rightarrow b = ae$ and $\frac{{2{b^2}}}{a} = 8$

$a\left( {1 - {e^2}} \right) = 4,{a^2}{e^2} = {a^2}\left( {1 - {e^2}} \right)$

$ \Rightarrow {e^2} = \frac{1}{2}$

$ \Rightarrow a = 8,b = 4\sqrt 2 $

Hence equation oh ellipse is $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{32}} = 1$

$\left( {4\sqrt 3 ,2\sqrt 2 } \right)$ lies on this

Let the length of the latus rectum of an ellipse with its major axis long $x -$ axis and center at the origin, be $8$. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?

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