Let S(3,4) and S (9,12) be two foci of an ellipse. If foot of the perpendicular from focus S to a tangent of the ellipse is (1,-4), then find the eccentricity of the ellipse.

Explanation

Let the line lx+my+n=0 be normal to the ellipse at the point `(a cos theta, b sin theta)`.
Now, the equation of the normal at `(a cos theta, b sin theta)` is
`(ax)/(cos theta)-(bx)/(sin theta)=a^(2)-b^(2)" "(1)`
Comparing this line with the line lx+my+n=0, we have
`(a)/(lcostheta)=(-b)/(m sin theta)=(a^(2)-b^(2))/(-h)`
`or costheta=(-an)/(l(a^(2)-b^(2)))`
and `sin theta=(bn)/(m(a^(2)-b^(2)))`
Squaring and adding, we get
`(a^(2)n^(2))/(l^(2)(a^(2)-b^(2))^(2))+(b^(2)n^(2))/(m^(2)(a^(2)-b^(2))^(2))=1`
`or (a^(2))/(l^(2))+(b^(2))/(m)=((a^(2)-b^(2)))/(n^(2))`