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  • If two points are taken on the minor axis of an ellipse, whose semi-major axis is a, at the same distance from the centre as the foci, then the sum of the squares of the perpendiculars from these p

If two points are taken on the minor axis of an ellipse, whose semi-major axis is a, at the same distance from the centre as the foci, then the sum of the squares of the perpendiculars from these points on any tangent to the ellipse is .

Option: 1

\mathrm{a^{2}}


Option: 2

\mathrm{\frac{a^{2}}{2}}


Option: 3

\mathrm{2a^{2}}


Option: 4

\mathrm{4a^{2}}


Answers (1)

best_answer

Let the equation of the ellipse be \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}.

Then, the distance of a focus from the centre

\mathrm{ =a e=a \sqrt{1-b^2 / a^2}=\sqrt{a^2-b^2} }
so that the two points on minor axis are

\mathrm{ \mathrm{S}_1\left(0, \sqrt{\mathrm{a}^2-\mathrm{b}^2}\right) \text { and } \mathrm{S}_1\left(0,-\sqrt{\mathrm{a}^2-\mathrm{b}^2}\right) \text {. } }

Now any tangent to the ellipse is \mathrm{y=m x+\sqrt{a^2 m^2+b^2}}, where m is a parameter.
The sum of the squares of the perpendiculars on this tangent from the two points \mathrm{S_1} and \mathrm{S_1{ }^{\prime}}

\mathrm{ \begin{aligned} & \text { is }\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 \\\\ & \left.=2\left(a^2-b^2+a^2 m^2+b^2\right) /\left(1+m^2\right)=2 a^2 \text { (constant }\right) . \end{aligned} }

Posted by

sudhir.kumar

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