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Focal Chord

Let a, r, s...

Question

Let $a, r, s, t$ be non-zero real numbers. Let $P (a t^{2}, 2 a t), Q, R (a r^{2}, 2 a r)$ and $S (a s^{2}, 2 a s)$ be distinct point on the parabola $y^{2} = 4 a x$ . Suppose that $P Q$ is the focal chord and lines $Q R$ and $P K$ are parallel, where $K$ is point $(2 a, 0)$
The value of $r$ is

A

- \frac{1}{t}

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B

\frac{t^{2} + 1}{t}

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C

\frac{1}{t}

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D

\frac{t^{2} - 1}{t}

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Solution

The correct option is C $\frac{t^{2} + 1}{t}$
Given that,

Slope of $(Q R) =$ Slope of $(P K)$

$\frac{2 a t - 0}{a t^{2} - 2 a} = \frac{- \frac{2 a}{t} - 2 a r}{\frac{a}{t^{2}} - a r^{2}}$

$\frac{2 a t}{a (t^{2} - 2)} = \frac{\frac{- 2 a - 2 a r t}{t}}{\frac{a - a r^{2} t}{t^{2}}} = \frac{t^{2} (- 2 a - 2 a r t)}{t (a - a r^{2} t^{2})} = \frac{- 2 a t (1 + r t)}{a (1 - r^{2} t^{2})}$

$\frac{2 t}{(t^{2} - 2)} = \frac{- 2 t (1 + r t)}{(1 - r^{2} t^{2})}$

$\frac{1}{(t^{2} - 2)} = \frac{(1 - r t)}{(1 - r^{2} t^{2})} = \frac{(1 + r t)}{(1 - r t) (1 + r t)} = \frac{1}{(1 - r t)}$

$\frac{1}{(t^{2} - 2)} = \frac{1}{(r t - 1)}$

$r t - 1 = t^{2} - 2$

$r t = t^{2} + 1$

$r = \frac{t^{2} + 1}{t}$

This is the answer.

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