If the eccentric angles of two points P and Q on the  ellipse  x2a2+y2b2  are  α,β  such that  α+β=π2 then,the locus of the point of intersection of the normals at  P  and  Q is

Equations of the normal at P  and Q  areax sec ⁡α−by cosec⁡α=a2−b2axsec⁡β− by cosec⁡β=a2−b2If they intersect at  (h, k). ahsec⁡α−bkcosec⁡α=ahcosec⁡α−bksec⁡α(∵β=π/2−α) (ah+bk)(sec⁡α−cosec⁡α)=0⇒ah+bk=0Locus of  (h,k) is ax+by=0