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  • If the normal at an end of latus rectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given byOption: 1</

If the normal at an end of latus rectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by

Option: 1

\mathrm{e^{4}+e^{2}-1=0}


Option: 2

\mathrm{ e^{2}+e-5=0}


Option: 3

\mathrm{\mathrm{e}^{3}=\sqrt{5}}


Option: 4

None of these


Answers (1)

best_answer

The equation of normal at \mathrm{\left(a e, b \sqrt{1-e^{2}}\right)\, is \, \frac{x-a e}{\frac{a e}{a^{2}}}=\frac{y-b \sqrt{1-e^{2}}}{\left(\frac{b \sqrt{1-e^{2}}}{b^{2}}\right)}}
Since it passes through \mathrm{(0,-b)}
\mathrm{\frac{-a e}{\left(\frac{a e}{a^{2}}\right)}=\frac{-b-b \sqrt{1-e^{2}}}{\left(\frac{b \sqrt{1-e^{2}}}{b^{2}}\right)}}

Putting \mathrm{b^{2}=a^{2}\left(1-e^{2}\right)} and simplifying

We get \mathrm{e^{4}+e^{2}-1=0}

Posted by

Rishabh

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