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Position of a Point with Respect to Circle

If the normal...

Question

If the normal at two points $P$ and $Q$ of a parabola $y^{2} = 4 a x$ intersect at a third point $R$ on the curve, then the product of ordinates of $P$ and $Q$ is

A

4 a^{2}

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B

2 a^{2}

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C

- 4 a^{2}

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D

8 a^{2}

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Solution

The correct option is D $8 a^{2}$
Let $(a t^{2}, 2 a t_{1}), (a t_{2}^{2}, 2 a t_{2}), (a t_{3}^{2}, 2 a t_{3})$ be coordinates of the point P, Q, R respecting which lie on the curve $y^{2} = 4 a x$ .

Now, equation of the normal to the curve at P is,

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

R is a point which lies at the normal at P.

$∴$ Substituting $y = 2 a t_{3}$ , $x = a t_{3}^{2}$ in equation $(1)$

$2 a t_{3} - 2 a t_{1} = - t_{1} (a t_{3}^{2} - a t_{1}^{2})$

$2 a (t_{3} - t_{1}) = - a t_{1} (t_{3}^{2} - t_{1}^{2})$

$2 (t_{3} - t_{1}) = - t_{1} (t_{3} - t_{1}) (t_{3} + t_{1})$

$2 = - t_{1} (t_{3} + t_{1})$

$\Rightarrow 2 = - t_{1} t_{3} - t_{1}^{2}$ ……….(ii)

Now, the equation of the normal to the curve at Q is

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

R is a point which lies at the normal at Q.

Substituting $y = 2 a t_{3}, x = a t_{3}^{2}$ in equation (iii)

$2 a t_{3} - 2 a t_{2} = - t_{2} (a t_{3}^{2} - a t_{2}^{2})$

$2 a (t_{3} - t_{2}) = - a t_{2} (t_{3} - t_{2}) (t_{3} + t_{2})$

$2 = - t_{2} (t_{3} + t_{2})$

$\Rightarrow 2 = - t_{2} t_{3} - t_{2}^{2}$ ……..(iv)

Multiplying equation (ii) with $t_{2}$ ,

$2 t_{2} = - t_{1} t_{2} t_{3} - t_{1}^{2} t_{2}$ ………..(v)

Multiplying equation (iv) with $t_{1}$ ,

$2 t_{1} = - t_{1} t_{2} t_{3} - t_{2}^{2} t_{1}$ …….(vi)

Subtracting (v) and (vi), we get

$2 t_{2} - 2 t_{1} = - t_{1}^{2} t_{2} + t_{2}^{2} t_{1}$

$2 (t_{2} - t_{1}) = - t_{1} t_{2} (t_{1} - t_{2})$

$2 (t_{2} - t_{1}) = t_{1} t_{2} (t_{2} - t_{1})$

$\Rightarrow t_{1} t_{2} = 2$

Now, the product of ordinates of P and Q $= 2 a t_{1} \cdot 2 a t_{2}$

$= 4 a^{2} t_{1} t_{2}$

$= 4 a^{2} (2)$

$= 8 a^{2}$ .

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