TP and TQ are tangents to the parabola and the normals at P and Q meet at a point R on the curve ; prove that the centre of the circle circumscribing the triangle TPQ lies on the parabola
$2 y^{2} = a (x - a) .$

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Solution

let the point of contact of tangents be $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$
Let centre of the circle through triangle $T P Q$ be $(h, k)$
Then point of intersection of tangents is $T (a t_{1} t_{2}, a (t_{1} + t_{2}))$
Equation of chord $P Q$ is
$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2} 2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2} = 0$
Equation of circle through $T P Q$ is
$(x - a t_{1}^{2}) (x - a t_{2}^{2}) + (y - 2 a t_{1}) (y - 2 a t_{2}) + λ (2 x - (t_{1} + t_{2}) y + 2 a t_{1} t_{2}) = 0..... (i)$
It passes through $T$
$(a t_{1} t_{2} - a t_{1}^{2}) (a t_{1} t_{2} - a t_{2}^{2}) + (a (t_{1} + t_{2}) - 2 a t_{1}) (a (t_{1} + t_{2}) - 2 a t_{2}) + λ (2 a t_{1} t_{2} - (t_{1} + t_{2}) a (t_{1} + t_{2}) + 2 a t_{1} t_{2}) = 0 \Rightarrow λ = - a (t_{1} t_{2} + 1)$
Simplifying $(i)$
$x^{2} + y^{2} - x (a t_{1}^{2} + a t_{2}^{2} - 2 λ) - y (2 a t_{1} + 2 a t_{2} + (t_{1} + t_{2}) λ) + a^{2} t_{1}^{2} t_{2}^{2} + 4 a^{2} t_{1} t_{2} + 2 a t_{1} t_{2} λ = 0$
Comparing the centre of the circle $(h, k)$ with general equation of circle
$\Rightarrow 2 h = a t_{1}^{2} + a t_{2}^{2} - 2 λ$
Substituting $λ$
$2 h = a t_{1}^{2} + a t_{2}^{2} - 2 (- a (t_{1} t_{2} + 1)) 2 h = a t_{1}^{2} + a t_{2}^{2} + 2 a t_{1} t_{2} + 2 a 2 h = a {{(t}_{1} + t_{2})}^{2} + 2 a . . . . . . (i i)$
Also,
$2 k = 2 a t_{1} + 2 a t_{2} + (t_{1} + t_{2}) λ 2 k = 2 a t_{1} + 2 a t_{2} + (t_{1} + t_{2}) (- a (t_{1} t_{2} + 1)) 2 k = a {2 (t_{1} + t_{2}) - 3 (t_{1} + t_{2})} 2 k = - a (t_{1} + t_{2}) t_{1} + t_{2} = - \frac{2 k}{a}$
Substituting in $(i)$
$2 h = a {(- \frac{2 k}{a})}^{2} + 2 a a h = 2 k^{2} + a^{2} 2 k^{2} = a (h - a)$
generalising the equation
$2 y^{2} = a (x - a)$
Hence proved

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