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  • The normal drawn to 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%5E2%7D%7Ba%5E2%7D&

The normal drawn to the ellipse   \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}   at the extremity of the latus rectum passes through the extremity of the minor axis. Eccentricity of this ellipse is equal to 

 

Option: 1

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


Option: 2

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


Option: 3

\sqrt{\frac{\sqrt{3}-1}{2}}


Option: 4

\frac{\sqrt{3}-1}{2}


Answers (1)

Equation of normal at \mathrm{ \left(a e, \frac{b^2}{a}\right)} , is
\mathrm{ \frac{x-a e}{\frac{a e}{a^2}}=\frac{y-\frac{b^2}{a}}{\frac{b^2}{a b^2}} \text { i.e., } \frac{a(x-a e)}{e}=\left(a y-b^2\right)}

It should pass through (0,-b)

\mathrm{ \Rightarrow \quad \frac{a(0-a e)}{e}}

\mathrm{ =-a b-b^2}
\mathrm{ \Rightarrow \quad a^2=a b+b^2 \Rightarrow \frac{b^2}{a^2}+\frac{b}{a}-1=0 \\ }

\mathrm{ \Rightarrow \quad \frac{b}{a}=\frac{-1+\sqrt{5}}{2} \Rightarrow b^2 \\ }

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

\mathrm{ 2 \Rightarrow e=\sqrt{\frac{\sqrt{5}-1}{2}} }.

Posted by

Kshitij

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