The normal at a point P of a parabola meets the curve again in Q, and T is the pole of PQ; show that T lies on the diameter passing through the other end of the focal chord passing through P, and that PT is bisected by the directrix.

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Solution

Let the point $P$ be $(a t^{2}, 2 a t)$ and point $Q$ be $(a t_{1}^{2}, 2 a t_{1})$

Equation of normal at $P$ is

$y = - t x + 2 a t + a t^{3} t x + y - 2 a t - a t^{3} = 0....... (i)$

Let the pole be $T (h, k)$

Equation of chord of contact is

$k y = 2 a x + 2 a h 2 a x - k y + 2 a h = 0........ (i i)$

Now $(i)$ and $(i i)$ represents the equation of same line

$\frac{2 a}{t} = \frac{- k}{1} = \frac{2 a h}{- 2 a t - a t^{3}} \Rightarrow h = - (2 a + a t^{2}) k = - \frac{2 a}{t}$

So the point $T$ is $(- (2 a + a t^{2}), - \frac{2 a}{t})$

For focal chord $t t_{1} = - 1$

$t_{1} = - \frac{1}{t}$

$\Rightarrow Q (\frac{a}{t^{2}}, - \frac{2 a}{t})$

Equation of diameter is $y = \frac{2 a}{m} . . . . . . . (i i i)$

$(i i i)$ passes through $Q$

$\Rightarrow m = - t$

So the equation of diameter is

$y = - \frac{2 a}{t}$

Clearly $Q$ lies on the equation of diameter.

Mid point of $P T$ is

$⎛ ⎜ ⎜ ⎜ ⎝ \frac{- 2 a - a t^{2} + a t^{2}}{2}, \frac{- \frac{2 a}{t} + 2 a t}{2} ⎞ ⎟ ⎟ ⎟ ⎠ (- a, a t - \frac{a}{t})$

Equation of directrix is $x = - a$

Clearly mid point of $T P$ lies on the directrix

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