Solution :
Let P(h,k) be the mid point of chord of the parabola \(y^2\) = 4ax,
so equation of chord is yk – 2a(x+h) = \(k^2\) – 4ah.
Since it passes through (p,q)
\(\therefore\) qk – 2a(p+h) = \(k^2\) – 4ah
\(\therefore\) Required locus is \(y^2\) – 2ax – qy + 2ap = 0
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