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y^2=4ax,2yy'=4a,y'=2a/y,slope of normal=-1/y' =-y/2a, equation of first normal=(y-y1)/(x-((y1)^2/4a))=-y1/2a,
2ay-2ay1=-y1x+((y1)3)/4a)... eq(1), 2ay-2ay2=-y2x+(((y2)^3)/4a) ...(2) solve the two previous simultaneous equations to obtain (X,Y) for the point of intersection. let Y^2=4aX, this will take you to the answer
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