Question
$P$ is a point on the parabola $y^{2} = 4x$ and $Q$ is a point on the line $2x + y + 4 = 0$. If the line $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then the co-ordinates of $P$ can be :
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Solution
Sol. [Key Concept: $x - y + 1 = 0$ being the perpendicular bisector of $PQ$ makes $P,Q$ mirror images in it; reflect a point of $2x + y + 4 = 0$ and intersect with the parabola.]
$Q$ is image of $P$ w.r.t. $x - y + 1 = 0$
$$\frac{x - t^{2}}{1} = \frac{y - 2t}{- 1} = - 2\left( \frac{t^{2} - 2t + 1}{1^{2} + 1^{2}} \right)$$
$$Q \equiv (x,y) = \left( 2t - 1,t^{2} + 1 \right)$$
Put $Q$ to equation $2x + y + 4 = 0$
$$\begin{matrix} \Rightarrow & 2(2t - 1) + \left( t^{2} + 1 \right) + 4 = 0 \\ & \ (t + 3)(t + 1) = 0 \Rightarrow t = - 1, - 3 \\ \Rightarrow & P \equiv (1, - 2),(9, - 6) \end{matrix}$$
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