1. The problem statement, all variables and given/known data If P and Q are two points on ellipse [(x^2/a^2)+(y^2/b^2)]=1 such that PQ subtends a right angle at the centre O then.Prove that 1/[(OP)^2] + 1/[(OQ)^2] = [1/(a^2)] +[1/(b^2)] 2. Relevant equations Parametric form of points P(acos(θ),bsin(θ)), 3. The attempt at a solution Since the angle between them is 90 degrees therefore the points are P[acos(θ),bsin(θ)] and Q[acos(90+θ),bsin(90+θ)] = Q[-asin(θ),bcos(θ)] FirstNow I tried the distance formula but in vain SecondI also tried using vectors as follows Position vector of P= acos(θ) i +bsin(θ) j Q= -asin(θ)i +bcos(θ) j Since OP and OQ are perpandicular their dot product is zero OP.OQ = 0 acos(θ).[-asin(θ)] + bsin(θ).[bcoz(θ)]=0 b^2[sin(θ)cos(θ)]- a^2[sin(θ)cos(θ) = 0 which gives a^2=b^2 a = b I have 2 questions, 1)How to prove this? 2)Why does the vector method give a=b (which i think is only a special case). Did I do anything wrong? Any help will be appreciated...Thanks.