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  • If the normal at one end of the latusrectum of the ellipse <img alt="\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx

If the normal at one end of the latusrectum of the ellipse \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1} passes through an end of the minor axis, then \mathrm{e=}

Option: 1

\frac{\sqrt{5}-1}{2}


Option: 2

\sqrt{\frac{\sqrt{5}-1}{2}}


Option: 3

\frac{5-\sqrt{5}}{2}


Option: 4

\sqrt{\frac{5-\sqrt{5}}{2}}


Answers (1)

best_answer

An end of the latus rectum is \left(\mathrm{ae}, \frac{\mathrm{b}^{2}}{\mathrm{a}}\right).

Normal at this point has equation

\mathrm{y-\frac{b^{2}}{a}=\frac{1}{e}(x-a e)}.  If it passes throught
\mathrm{(0,-b), then \: a^{2}-a b-b^{2}=0}

\mathrm{\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=\frac{\sqrt{5}-1}{2}=\sqrt{\mathrm{e}^{2}-1}}
\mathrm{\Rightarrow \mathrm{e}^{2}-1=\frac{3-\sqrt{5}}{2} \Rightarrow \mathrm{e}=\sqrt{\frac{5-\sqrt{5}}{2}}}

Posted by

manish painkra

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