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  • The normal at an end of a latus-rectum 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

The normal at an end of a latus-rectum of the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} passes through an end of the minor axis if

 

Option: 1

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


Option: 2

\mathrm{e^3+e^2=1}


Option: 3

\mathrm{e^2+e=1}


Option: 4

\mathrm{e^3+e=1}


Answers (1)

Let an end of a latus-rectum be \mathrm{\left(a e, b \sqrt{1-e^2}\right)} , then the equation of the normal at this end is

\mathrm{\frac{x-a e}{a e / a^2}=\frac{y-b \sqrt{1-e^2}}{b \sqrt{1-e^2} / b^2}}

It will pass through the end (0, −b) if 

\mathrm{\begin{aligned} & -a^2=\frac{-b^2\left(1+\sqrt{1-e^2}\right)}{\sqrt{1-e^2}} \quad \text { or } \frac{b^2}{a^2}=\frac{\sqrt{1-e^2}}{1+\sqrt{1-e^2}} \\ & \text { or } \quad\left(1-e^2\right)\left[1+\sqrt{1-e^2}\right]=\sqrt{1-e^2} \\ & \text { or } \quad \sqrt{1-e^2}+1-e^2=1 \text { or } e^4+e^2=1 . \end{aligned}}

 

Posted by

Ramraj Saini

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