Shivank Shekhar
Last Activity: 6 Years ago
For simplicity assume that that one of the normals drawn at the vertex of the parabola Y^2=4ax and other two are drawn at the points that is P t1 and q t2.
Now it means that normal will meet on the x axis.
A= 2a+ a(t1^2+t2^2+t1t2), -at1t2(t1+t2)
Since Y co-ordinate of A is zero therefore T1 is equal to T2
PS=a
QS=SR=a+at^2
SA= a+at^2
Therefore PS.QS.SR=aSA^2.