The locus of centres of a family of circles passing through the vertex of the parabola <img alt="\mathrm{y}^2=4 \mathrm{ax}" src="https://entrancecorner.oncodecogs.c

The locus of centres of a family of circles passing through the vertex of the parabola $\mathrm{y}^2=4 \mathrm{ax}$ and cutting the parabola orthogonally at the other point of intersection is $\mathrm{2 y^2\left(2 y^2+x^2-12 a x\right)=a x(3 x-4 a)^2 k^2}$ , where k =

Option: 1
$\pm 2$

Option: 2
$\pm \frac{1}{2}$

Option: 3
$\pm 1$

Option: 4
$\pm \sqrt{3}$

Answers (1)

Let $\mathrm{P\left(a t^2, 2 a t\right) be\, any\, point \, on\, y^2=4 a x. }$ $\mathrm{P\left(a t^2, 2 a t\right) be\, any\, point \, on \, y^2=4 a x. }$ Then vertex is A(0,0). The equation of tangent at P is
$\mathrm{\text { ty }=x+a t^2 }$ ...........................(1)
Tangent at $\mathrm{ \mathrm{P} }$ will be normal to the circle, AP is a chord whose mid point is $\mathrm{ \left(\frac{\mathrm{at}^2}{2}\right., at ) }$ and slope is $\mathrm{ \frac{2}{\mathrm{t}} }$
$\mathrm{ \therefore \quad }$ Equation of the line passing through mid-point of AP and perpendicular to $\mathrm{ \mathrm{AP}}$ is
$\mathrm{ \begin{aligned} & y-a t=-\frac{t}{2}\left(x-\frac{a t^2}{2}\right) \\ & t x+2 y=\frac{a t^3}{2}+2 a t \end{aligned} }$ ........................(2)
(1) and (2) both pass through $\mathrm{\left(\mathrm{x}_1, \mathrm{y}_1\right) }$ which is the centre of the circle
$\mathrm{\begin{aligned} & \mathrm{ty}_1=\mathrm{x}_1+\mathrm{at}^2.........(3) \\ & 2 \mathrm{tx}_1+4 \mathrm{y}_1=a t^3+4 \mathrm{at} \end{aligned}......(4) }$
Multiplying (3) by t and subtracting (4) we have
$\mathrm{t^2 y_1+t\left(4 a-3 x_1\right)-4 y_1=0 }$
Also from (3), $\mathrm{\mathrm{at}^2-\mathrm{ty}_1+\mathrm{x}_1=0......(5) }$
$\mathrm{Eliminating { }^t from(5) \& (6) }$
$\mathrm{\frac{t^2}{x_1\left(4 a-3 x_1\right)-4 y_1^2}=\frac{t}{-4 a y_1-x_1 y_1}=\frac{1}{-y_1^2-a\left(4 a-3 x_1\right)} }$
On simplifying we get
$\mathrm{2 \mathrm{y}_1^2\left(2 \mathrm{y}_1^2+\mathrm{x}_1^2-12 \mathrm{ax}_1\right)=\mathrm{ax}_1\left(3 \mathrm{x}_1-4 \mathrm{a}\right)^2}$
Hence required locus is
$\mathrm{2 y^2\left(2 y^2+x^2-12 a x\right)=a x(3 x-4 a)^2 k^2 .}$