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If two points are taken on the minor axi...

If two points are taken on the minor axis of an ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` at the same distance from the center as the foci, then prove that the sum of the squares of the perpendicular distances from these points on any tangent to the ellipse is `2a^2dot`

Text Solution

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Let the equationof the ellipse be
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`

Then the distane of a focus from the center is
`ae=asqrt(1-(b^(2))/(a^(2)))=sqrt(a^(2)-b^(2))`
so that the two points on the minor aixs are
`S_(1)(0,sqrt(a^(2)-b^(2)))and S_(1)'(0,-sqrt(a^(2)-b^(2)))`
Now, any tangent to the ellipse is
`y=mx+sqrt(a^(2)m^(2)+b^(2))`, where m is a parameter
The sum of the squaes of the perpendicular on this tangent from the two points `S_(1) and S_(1)'` is
`((sqrt(a^(2)-b^(2))-sqrt(a^(2)m^(2)+b^(2)))/(sqrt(1+m^(2))))^(2)+((-sqrt(a^(2)-b^(2))-sqrt(a^(2)m^(2)+b^(2)))/(sqrt(1+m^(2))))^(2)`
`=(2(a^(2)-b^(2)+a^(2)m^(2)+b^(2)))/(1+m^(2))=2a^(2)` (constant)
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