If from the vertex of a parabola a pair of chords be drawn at right angles to one another and with these chords as adjacent sides a rectangle be made, prove that the locus of the further angle of the rectangle is the parabola $y^{2} = 4 a (x - 8 a) .$

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Solution

Let the vertex $P$ be $(a t_{1}^{2}, 2 a t_{1})$ and vertex $Q$ be $(a t_{2}^{2}, 2 a t_{2})$

and let the further vertex be $R (h, k)$

Slope of $O P = \frac{2 a t_{1} - 0}{a t_{1}^{2} - 0} = \frac{2}{t_{1}}$

Similarly slope of $O Q = \frac{2}{t_{2}}$

As both the line are perpendicular

$∴ \frac{2}{t_{1}} \times \frac{2}{t_{2}} = - 1 t_{1} t_{2} = - 4.... (i)$

Mid point of $P Q$ is

$(\frac{a t_{1}^{2} + a t_{1}^{2}}{2}, \frac{2 a t_{1} + 2 a t_{2}}{2}) (\frac{a t_{1}^{2} + a t_{1}^{2}}{2}, a t_{1} + a t_{2})$

Mid point of $O R$ is

$(\frac{h}{2}, \frac{k}{2})$

As the diagonals of rectangle bisect each other

$∴ (\frac{a t_{1}^{2} + a t_{1}^{2}}{2}, a t_{1} + a t_{2}) = (\frac{h}{2}, \frac{k}{2}) h = a t_{1}^{2} + a t_{1}^{2} t_{1}^{2} + t_{1}^{2} = \frac{h}{a} . . . . . . . (i i) a t_{1} + a t_{2} = \frac{k}{2} t_{1} + t_{2} = \frac{k}{2 a}$

Squaring both sides

$t_{1}^{2} + t_{2}^{2} + 2 t_{1} t_{2} = \frac{k^{2}}{4 a^{2}}$

Substituting $(i)$ and $(i i)$

$\frac{h}{a} - 8 = \frac{k^{2}}{4 a^{2}} 4 a (h - 8 a) = k^{2} k^{2} = 4 a (h - 8 a)$

generalising the equation

$y^{2} = 4 a (x - 8 a)$

Hence proved

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