If the tangents at P and Q in the parabola meet in T, then which of the following statements are correct.

1. TP and TQ subtends equal angle at the focus S

2. $S T^{2}$ = SP.SQ

Only 1

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Only 2

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Both 1 2

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None of these

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Solution

The correct option is C
Both 1 2

Let's take a standard parabola $y^{2} = 4 a x$ and draw tangents

at $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ ... Which meet at T.

We know the point of intersection of tangent is $T (a t_{1} t_{2}, a (t_{1} + t_{2}))$

We can also calculate coordinates of T by calculating tangents

at P $&$ Q and their point of intersection is point T.

Tangent at P

$y (2 a t_{1}) = 2 a (x + a t_{1}^{2})$

$y t_{1} = x + a t_{1}^{2}$ ............(1)

Similarly tangent at Q

$y t_{2} = x + a t_{2}^{2}$ ............(2)

Solving equation (1) $&$ (2)

We get, $x = a t_{1} t_{2}$

$y = a (t_{1} + t_{2})$

Coordinates of $T (a t_{1} t_{2}, a (t_{1} + t_{2})$

Statement 1

To,check $α = β$ , or check that T lies on the angle bisector

of the $∠ P S Q$ . i.e., perpendicular distance of T from the

line SP is equal to the perpendicular of T from SQ.

Equation of SP

$y = \frac{2 a t_{2}}{a t_{1}^{2} - a} (x - a)$

$2 t_{1} x - (t_{1}^{2} - 1) y - 2 a t_{1} = 0$

$P_{1} = \frac{2 a t_{1}^{2} t_{2} - (t_{1}^{2} - 1) a (t_{1} + t_{2}) - 2 a t_{1}}{\sqrt{(t_{1}^{2} - 1)^{2} + 4 t_{1}^{2}}}$

$P_{1} = a | t_{1} - t_{2} |$

Similarly equation of SQ

$y = \frac{2 a t_{2}}{a t_{2}^{2} - a} (x - a)$

$2 t_{2} x - (t_{2}^{2} - 1) - 2 a t_{2} = 0$

Perpendicular distance of SQ from T

$P_{2} = \frac{2 a t_{2}^{2} t_{1} - (t_{2}^{2} - 1) a (t_{1} + t_{2}) - 2 a t_{2}}{\sqrt{(t_{2}^{2} - 1)^{2} + 4 t_{2}^{2}}}$

So, we can say that $α = β$

Statement 1 is correct.

Statement 2:

$S T^{2} = (a - a t_{1} t_{2})^{2} + (a (t_{1} + t_{2}) - 0)^{2}$

$= a^{2} (1 - t_{1} . t_{2})^{2} + a^{2} (t_{1} + t_{2})^{2}$

$= a^{2} [1 + t_{1}^{2} t_{2}^{2} + t_{1}^{2} + t_{2}^{2}]$

$S P = \sqrt{(a t_{1}^{2} - a)^{2} + (2 a t_{1} - 0)^{2}} = \sqrt{a^{2} (t_{1}^{2} - 1) + 4 a^{2} t_{1}^{2}}$

$= a \sqrt{t_{1}^{4} + 1 - 2 t_{1}^{2} + 4 t_{1}^{2}} = a \sqrt{(t_{1}^{2} + 1)^{2}} = a (t_{1}^{2} + 1)$

$S Q = \sqrt{(a t_{2}^{2} - a)^{2} + (2 a t_{1} - 0)^{2}} = \sqrt{a^{2} (t_{2}^{2} + 1) + 4 a^{2} t_{2}^{2}}$

$= a \sqrt{t_{2}^{4} + 1 - 2 t_{2}^{2} + 4 t_{2}^{2}} = a \sqrt{(t_{2}^{2} + 1)^{2}} = a (t_{2}^{2} + 1)$

$S P . S Q = a (t_{1}^{2} + 1) \times a (t_{2}^{2} + 1)$

$= a^{2} [t_{1}^{2} t_{2}^{2} + t_{1}^{2} + t_{2}^{2} + 1]$

$S P . S Q = a^{2} [1 + t_{1}^{2} t_{2}^{2} + t_{1}^{2} + t_{2}^{2}] = S T^{2}$

Statement 2 is correct

Both the statement are correct.

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