PQ is chord of contact of tangents from point T to a parabola. If PQ is normal at P, then the directrix divides PT in the ratio

1 : 2

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2 : 1

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3 : 1

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1 : 1

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Solution

The correct option is D $1 : 1$
Consider a standard parabola $y^{2} = 4 a x$
Let $P = (a t_{1}^{2}, 2 a t_{1})$ and $Q = (a t_{2}^{2}, 2 a t_{2})$ be the two points.

Equation of tangent at P is $t_{1} y = x + a t_{1}^{2}$ ...[1]
Equation of tangent at Q is $t_{2} y = x + a t_{2}^{2}$ ...[2]
These two lines intersect at the point $T$
Solving [1] and [2], we get
$T = (a t_{1} t_{2}, a (t_{1} + t_{2}))$

PQ is normal at P
Therefore, slope of normal at P = slope of PQ
$∴ - t_{1} = \frac{2 a t_{1} - 2 a t_{2}}{a t_{1}^{2} - a t_{2}^{2}}$
$∴ t_{2} = - t_{1} - \frac{2}{t_{2}}$
So, co-ordinate of $T = (a t_{1} t_{2}, a (t_{1} + t_{2})) = (a t_{1} (- t_{1} - \frac{2}{t_{2}}), a (t_{1} + (- t_{1} - \frac{2}{t_{2}})))$
$∴ T = (- a t_{1}^{2} - 2 a, - \frac{2 a}{t_{1}})$
$P = (a t_{1}^{2}, 2 a t_{1})$
Midpoint of PT $= (- a, a (t_{1} - \frac{1}{t_{1}}))$ ...[3]

Equation of PT is $t_{1} y = x + a t_{1}^{2}$
Equation of directrix is $x = - a$
The directrix and PT intersect at the point $(- a, a (t_{1} - \frac{1}{t_{1}}))$ ...[4]
From [3] and [4] directrix cuts the segment PT at its midpoint.
So, directrix divides PT in the ratio 1:1

So, the answer is option (D).

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