A circle cuts the parabola y2= 4ax in the points ( ati2, 2ati) for i = 1, 2, 3, 4. Prove that ti+ t2+ t3+ t4= O.

We can proceed in the following way

Let the equation of the circle be x^2+y^2+2*g*x+2*f*y+c=0

Let the equation of the parabola be y^2=4ax.

Now replace x^2 by y^4/16a^2 in the equation of circle and x by y^2/4a.(To find the ordinates of the intersection points).

Now the equation of the circle becomes.

y^4/16*a^2+y^2+2*g*(y^2/4a)+2*f*y+c=0 (The roots of the equation give the value of the ordinates of the point of intersection)

The co-efficient of y^3 is 0. … (In general).(The roots may be real or complex).

So Sum of ordinates is 0.