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  • There are exactly two points on the ellipse <img alt="\mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm

There are exactly two points on the ellipse \mathrm{\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1} whose distance from its centre is same and is equal to\mathrm{\sqrt{\frac{a^2+2 b^2}{2}} .} Find the eccentricity of the ellipse.

Option: 1

\frac{1}{2}


Option: 2

\mathrm{\frac{1}{\sqrt{2}}}


Option: 3

\frac{1}{\sqrt{3}}


Option: 4

\sqrt{\frac{2}{3}}


Answers (1)

best_answer

Since there are exactly two points on the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, whose distance from centre is same, the points would be either end points of the major axis or of the minor axis.

But \mathrm{\sqrt{\frac{a^2+2 b^2}{2}}>b}, so the points are the vertices of major axis.

Hence \mathrm{a=\sqrt{\frac{a^2+2 b^2}{2}} \Rightarrow a^2=2 b^2}

Therefore \mathrm{\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\frac{1}{\sqrt{2}}}

Posted by

Sanket Gandhi

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