If the normal at an end of a latus rectum of | vedclass

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Question

$\frac{ a ^{2} x }{ x _{1}}-\frac{ b ^{2} y }{ y _{1}}= a ^{2} e ^{2}$

$\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2}} \cdot a=a^{2} e^{2}$

$\frac{ a

Vedclass · Accepted Answer

$e ^{4}+ e ^{2}-1=0$

Vedclass · Answer

$\frac{ a ^{2} x }{ x _{1}}-\frac{ b ^{2} y }{ y _{1}}= a ^{2} e ^{2}$

$\frac{a^{2} x}{a e}-\frac{b^{2} y}{b^{2}} \cdot a=a^{2} e^{2}$

$\frac{ ax }{ e }- ay = a ^{2} e ^{2} \Rightarrow \frac{ x }{ e }- y = ae ^{2}$

passes through $(0,$ b) $-b=a e^{2} \Rightarrow b^{2}=a^{2} e^{4}$

$a^{2}\left(1-e^{2}\right)=a^{2} e^{4} \Rightarrow e^{4}+e^{2}=1$

If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies

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