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  • If two points are taken on minor axis of an ellipse<img alt="\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20%5Cfrac%7Bx%5E2%7D%7Ba%

If two points are taken on minor axis of an ellipse\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at the same distance from the centre and the foci, the sum of the squares of the perpendiculars from these points on any tangent to the ellipse if a>b is

Option: 1

\mathrm{a^2}


Option: 2

\mathrm{b^2}


Option: 3

\mathrm{2a^2}


Option: 4

\mathrm{2b^2}


Answers (1)

\mathrm{\text { Given } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

OS = ae

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

\mathrm{\text { So two points on the minor axis are } S_1\left(0, \sqrt{a^2-b^2}\right) \text { and }S_1^{\prime}\left(0,-\sqrt{a^2-b^2}\right)}

\mathrm{\text { Let tangent to the ellipse be } y=m x+c=m x+\sqrt{a^2 m^2+b^2} \text {, }}

where m is a parameter.
Now sum of the squares of perpendiculars on this tangent
from the points S1 & S1′ is

\mathrm{\begin{aligned} & \left(\frac{\sqrt{a^2-b^2}-\sqrt{a^2 m^2+b^2}}{\sqrt{1+m^2}}\right)^2+\left(\frac{-\sqrt{a^2-b^2}-\sqrt{a^2 m^2+b^2}}{\sqrt{1+m^2}}\right)^2 \\ & =2\left(\frac{a^2-b^2+a^2 m^2+b^2}{1+m^2}\right)=\frac{2 a^2\left(1+m^2\right)}{1+m^2}=2 a^2 \end{aligned}}

Posted by

Kshitij

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