If the length of the latus rectum of an ellip | vedclass

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Question

Let for ellipse coordinates of focus and vertex are $(ae,0)$ and $(a,0)$ respectively.

$	herefore $ Distance between focus and vertex

$ = a\lef

Vedclass · Accepted Answer

$\frac {1}{3}$

Vedclass · Answer

Let for ellipse coordinates of focus and vertex are $(ae,0)$ and $(a,0)$ respectively.

$	herefore $ Distance between focus and vertex

$ = a\left( {1 - e} ight) = \frac{3}{2}$                       (given)

$ \Rightarrow a - \frac{3}{2} = ae$

$ \Rightarrow {a^2} + \frac{9}{4} - 3a = {a^2}{e^2}\,\,\,\,\,\,\,......\left( i ight)$

Length of latus rectum $ = \frac{{2{b^2}}}{a} = 4$

$ \Rightarrow {b^2} = 2a\,\,\,\,\,\,\,\,\,\,......\left( {ii} ight)$

         ${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}$

$ \Rightarrow {e^2} = 1 - \frac{{2a}}{{{a^2}}}$         (from$(ii)$)

$ \Rightarrow {e^2} = 1 - \frac{2}{a}\,\,\,\,\,\,\,\,\,.......\left( 3 ight)$

Substituting the value of ${e^2}$ in eq. $(i)$ we get;

$ \Rightarrow {a_2} + \frac{9}{4} - 3a = {a^2}\left( {1 - \frac{2}{a}} ight)$

$ \Rightarrow a = \frac{9}{4}$

$	herefore $ from eq. $(iii)$ we get;

${e^2} = 1 - \frac{2}{a} = 1 - \frac{8}{9} = \frac{1}{9}$

$ \Rightarrow e = \frac{1}{3}$

If the length of the latus rectum of an ellipse is $4\,units$ and the distance between a focus and its nearest vertex on the major axis is $\frac {3}{2}\,units$ , then its eccentricity is?

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