If a chord PQ of the parabola y^2=4ax subtends a right angle at the vertex, show that the locus of point of intersection of the normals at P and Q is y^2=16a^2(x-6a).

Hi,

Let point p & q BE IN PARAMETRIC FORM REPRESENTED BY PARAMETRIC FACTOR T1 AND T2

Now these two points substends right angles at origin so find out the slopes of PO and QO and put the multiplication of slopes equal to -1 (where O is origin)

This will give you one relation on t1 and t2.

Wirte the equation of normals from t1 and t2 ie from P and Q.
Find the intersection point in terms of t1 and t2.

Now substitute t1 in terms of t2 or vice versa from the condition we got from slope.

Now you have your intersection point (x, y coordinates) in terms of t1/t2.
Eliminate the lone t variable to get the answer.

Approved