IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

raulrag009

Let the points be (asecA,btanA) and (asecB,btanB)

Eqn of normals will be

axcosA+bycotA=a²+b²

and

axcosB+bycotB=a²+b²

coordinates of P will be y = (a²+b²)/bcotA

coordinates of Q will be y = (a²+b²)/bcotB

CP.CQ = (a²+b²)² / b²cotAcotB

As A+B = pie/2

Takin cot on both sides

cot(A+B)=cotpie/2

cotAcotB-1 =0

cotAcotB =1

CP.CQ = (a²+b²⁾²/ b^2

As b^2=a^2(e²-1) and a^2+b^2 = a^2e^2

therefore

CP.CQ = a^4e^4 / a^2(e^2-1)

CP.CQ=(a^2e^4)/e^2-1