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  • If the maximum distance of normal to the ellipse  <img alt="\frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1\, b< 2" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cfrac%7Bx%5E%7B2%7D%7D%

If the maximum distance of normal to the ellipse  \frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1\, b< 2 ,  from the origin is 1 , then the eccentricity of the ellipse is :

Option: 1

\frac{1}{2}


Option: 2

\frac{\sqrt3}{4}


Option: 3

\frac{\sqrt3}{2}


Option: 4

\frac{1}{\sqrt2}


Answers (1)

best_answer

Normal to the ellipse  \frac{x^{2}}{4}+\frac{y^{2}}{b^{2}}=1\, b< 2    at point (a cos \theta , b sin \theta) is    \frac{ax}{\cos \theta}- \frac{by}{\sin \theta}=a^{2}-b^{2}

Its distance from origin is

\begin{aligned} & \mathrm{d}=\frac{\left|\mathrm{a}^2-\mathrm{b}^2\right|}{\sqrt{\mathrm{a}^2 \sec ^2 \theta+\mathrm{b}^2 \operatorname{cosec}^2 \theta}} \\ & \mathrm{d}=\frac{\left|\mathrm{a}^2-\mathrm{b}^2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+2 a b+(a \tan \theta-\mathrm{b} \cot \theta)^2}} \end{aligned}

\begin{aligned} & \mathrm{d} \frac{|(a-b)(a+b)|}{\sqrt{a^2+b^2+2 a b+(a \tan \theta-b \tan \theta)^2}} \\ & d_{\max }=\frac{|(a-b)(a+b)|}{a+b}=|a-b| \\ & \because d_{\max }=1 \\ & |2-\mathrm{b}|=1 \\ & 2-\mathrm{b}=1[\because \mathrm{b}<2] \end{aligned}

b =1 

\begin{aligned} & \text { Eccentricity }=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} \\ & \Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2} \end{aligned}


 

Posted by

Ajit Kumar Dubey

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