Parabola ; part 3 ques?

Abhinav i m doing the problem for y^2=4ax. We can get the answer to ur problem by making appropriate replacements at the end

let t1 and t2 be the parameters of P and Q respectively. Then as OP and OQ are perpendicular, we get the relation t1t2=-4. Now let the coordinates of point R be (x,y)

Now use the property of rectangle the diagonals bisect each other, so we get two relations

at1^2+at2^2=x

and 2at1+2at2=y

now get the values of the redundant of these three equations is the req sol.

Use the identity (t1+t2)^2=t1^2+t2^2+2t1t2

At the last to get ur answer put a=1/4 and replace x by y and y by x

since y=x^2 so take paramateric as (x,x^2) at P and (x1,x1^2) at Q

after that OP and OQ are perpendicular, so x*x1=-1 -------------------(1)

then use the property of rectangle that the diagonals bisect each other

so, x-1/x = h ---------(2)

x^2+1/x^2 = K ----------------(3) where (h,k) is point at R

square the equation (2) both side and replace it with equation (3)

you get the answer h^2=k-2