2y 2 (2y 2 + x 2 – 12ax) = ax(3x – 4a) 2

Sol. Let P(at 2 , 2at) be any point on y 2 = 4ax. Then vertex A(0, 0). The equation of tangent at P is

ty = x + at 2 ...(i)

Tangent at P will be normal to the circle, AP is a chord whose mid point is and slope is

∴ Equation of the line passing through mid point of AP and perpendicular to AP is

y – at =

tx + 2y = + 2at ...(ii)

(i) and (ii) both pass through (x 1 , y 1 ) which is the centre of the circle

ty 1 = x 1 + at 2 ...(iii)

2tx 1 + 4y 1 = at 3 + 4at ...(iv)

Multiplying (iii) by t and subtracting (iv), we have

t 2 y 1 + t(4a – 3x 1 ) – 4y 1 = 0 ...(v)

also from (iii),

at 2 – ty 1 + x 1 = 0 ...(vi)

Eliminating t from (v) and (vi)

on simplyfying, we get

= ax 1 (3x 1 – 4a) 2

Hence required locus is 2y 2 (2y 2 + x 2 – 12ax) = ax(3x – 4a) 2

──────────────────────────────────────────────────────────────────────────────────────────