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Point Form of Tangent: Hyperbola

On the axis o...

Question

On the axis of any parabola $y^{2} = 4 a x$ there is a certain point $K$ on the X - axis which has the property that, if a chord $P Q$ of the parabola be drawn through it, then $\frac{1}{(P K)^{2}} + \frac{1}{(Q K)^{2}}$ is same for all positions of the chord. Find the coordinates of the point $K$ .

(4 a, 0)

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(2 a, 0)

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(a, 0)

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none of these

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Solution

The correct option is A $(2 a, 0)$
Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$
on $y^{2} = 4 a x$
equation of chord of PQ
$(t_{1} + t_{2}) y = 2 x + 2 a t_{1} t_{2}$ ....(1)
Point on x-axis is $K (- a t_{1} t_{2}, 0)$
$P K^{2} = (a t_{1}^{2} + a t_{1} t_{2})^{2} + 4 a^{2} t_{1}^{2}$
$= a^{2} t_{1}^{2} ((t_{1} + t_{2})^{2} + 4)$
$Q K^{2} = a^{2} t_{2}^{2} ((t_{1} + t_{2}) + 4)$
$\frac{1}{P K^{2}} + \frac{1}{Q K^{2}} = \frac{1}{a^{2} t_{1}^{2} ((t_{1} + t_{2})^{2} + 4)} + \frac{1}{a^{2} t_{2}^{2} ((t_{1} + t_{2})^{2} + 4)}$
$= \frac{t_{2}^{2} + t_{1}^{2}}{a^{2} t_{1}^{2} t_{2}^{2} ((t_{1} + t_{2})^{2} + 4)}$
$= \frac{t_{1}^{2} + t_{2}^{2}}{a^{2} t_{1}^{2} t_{2}^{2} ((t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} + 4)}$
$\frac{1}{P K^{2}} + \frac{1}{Q K^{2}}$ will be independent of K
$\Rightarrow \frac{t_{1}^{2} + t_{2}^{2}}{a^{2} t_{1}^{2} t_{2}^{2} ((t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} + 4)} \Rightarrow t_{1} t_{2} = - 2$
so fixed point $K (2 a, 0)$

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