Prove that the locus of the centre of the circle, which passes through the vertex of a parabola and through its intersections with a normal chord, is the parabola $2 y^{2} = a x - a^{2}$ .

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Solution

Equation of normal at point $(a t_{1}^{2}, 2 a t_{1})$ of the parabola $y^{2} = 4 a x$ is $y = - t_{1} x + 2 a t_{1} +$ , this line intersects the parabola at point $(a t_{2}^{2}, 2 a t_{2})$ , such that $t_{2} + t_{1} = - 2 / t_{1}$
The equation of any circle passing through $(0, 0)$ is
$x^{2} + y^{2} - 2 g x - 2 f y = 0$ .
If this passes through $(a t_{1}^{2}, 2 a t_{1})$ . then
$a t_{1}^{3} + 4 a t_{1} - 2 g t_{1} - 4 f = 0$ ...(2)
Similarly, if the circle passes through $(a t_{2}^{2}, 2 a t_{2})$ , then we have
$a t_{2}^{3} + 4 a t_{1} - 2 g t_{2} - 4 f = 0$ ...(3)
The locus of the centre $(g, f)$ of the circel will be obtined by eliminating $t_{1}, t_{2}$ and $t_{3}$ from (1), (2) and (3).
Subtracting (3) from (2), we have
$a (t_{1}^{3} - t_{2}^{3}) + 4 a (t_{1} - t_{2}) - 2 g (t_{1} t_{2}) = 0$
$\Rightarrow a (t_{1}^{2} + t_{2}^{3} + t_{1} t_{2}) + 4 a - 2 g = 0$
$\Rightarrow a (t_{2}^{2} - 2) + 4 a - 2 g = 0$ [by (1)]
$\Rightarrow a t_{2}^{2} + 2 a - 2 g = 0$ . ...(4)
Again multiplying (2) by $t_{2}$ and (3) by $t_{1}$ and then subtracting, we have
$a t_{1} t_{2} (t_{1}^{2} - t_{2}^{2}) + 4 f (t_{1} - t_{2}) = 0$
$\Rightarrow a (t_{1} t_{2} (t_{1} + t_{2}) + 4 f = 0$ [by (1)]
$\Rightarrow - 2 a t + 4 f = 0$
Hence $t_{2} = - 2 f / a$ .
Substituting this value of $t_{2}$ in (4), we have
$a (4 f^{2} / a^{2}) + 2 a - 2 g = 0$ or $2 f^{2} + a^{2} - a g = 0$ .
Gemeralising, the lous of the centre $(g, f)$ is
$2 y^{2} + a^{2} - a x = 0$
$2 y^{2} = a x - a^{2}$

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