If the end points $P (t_{1})$ and $Q (t_{2})$ of chord of parabola $y^{2} = 4 a x$ satisfy the relation $t_{1} t_{2} = k (c o n s t a n t)$ then prove that the chord always through a fixed point. Find that point also?

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Solution

Let $P (a t_{1}^{2}, 2 a t_{1})$ and $Q (a t_{2}^{2}, 2 a t_{2})$ be the
end pts is of a chords.

$∴$ equation of that chord be

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

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

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

$∴ y (t_{1} + t_{2}) - 2 a (k) = 2 x$

$[∵ t_{1} t_{2} = k]$

This chord always passes through a fixed pt $(- a k, 0)$ as

$0 (t_{1} + t_{2}) - 2 a k = 2 (- a k)$

$0 - 2 a k = - 2 a k$

$∴$ that pt is $(- a k, 0)$

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If the end points P(t1) and Q(t2) of chord of parabola y2=4ax satisfy the relation t1t2=k(constant) then prove that the chord always through a fixed point. Find that point also?

If the end points $P (t_{1})$ and $Q (t_{2})$ of chord of parabola $y^{2} = 4 a x$ satisfy the relation $t_{1} t_{2} = k (c o n s t a n t)$ then prove that the chord always through a fixed point. Find that point also?