Show that feet of the perpendicular from the foci upon any tangent to the ellipse lie on auxiliary circle of the ellipse.

We know that equation of any tangent to ellipse is y=mx+(a^2m^2+b^2)^1/2 ..........(1)Any point F (h,k) lies on tangent k-mh=(a^2m^2+b^2)^1/2 .............(2)Focus S (ae,0) slope of line SF = (k-0)/h-aeSF is perpendicular to line(1) m1m2=-1 k-0/h-ae*m =-1 》m=ae-h/k 》km+h=ae... (4)Square individually 3 and 4 equation and addK^2+m^2+k^2m^2+h^2=a^2m^2+b^2+a^2e^2 》k^2 (1+m^2)+h^2 (1+m^2)=a^2 (1+m^2)》h^2+k^2=a^2》therefore locus is x^2+y^2=a^2Which is equation of auxiliary circle