Solve this on ellipse ?

let the first tangent touch the ellipse at point (xP, yP),

and the perpendicular tangent touch the ellipse in point (xP', yP')

the tangent at point (xP, yP) of the ellipse has the equation

x*xP/a^2 + y*yP/b^2 = 1 --- > rewrite to

y = - xP*b^2/(yP*a^2)*x + b^2/yP

It intersects with the circle

x^2 + y^2 = a^2 + b^2 around the ellipse in point (xM, yM).

This circle is the location of all points from which the two possible tangents to the ellipse are perpendicular to each other. Its name is director's circle.

http://www.math.uoc.gr/~pamfilos/eGallery/problems...

Tthe tangent perpendicular to the first tangent has slope

m' = +yP*a^2/(xP*b^2).

and its equation is

y = yP*a^2/(xP*b^2)*x + B,

it goes through (xM, yM) on the master circle, and touches the ellipse in the point (xP', yP')

B = yM - yP*a^2/(xP*b^2)*xM

so its equation is

y = yP*a^2/(xP*b^2)*x + yM - yP*a^2/(xP*b^2)*xM

you find (xM, yM) by the usual procedure to find intersection points between line and circle.

and you find point (xP', yP') by setting the tangent equation and the ellipse equation equal.

OG