Equation of a line is given by $y + 2 a t = t (x - a t^{2})$ , $t$ being the parameter. Find the locus of the point of intersection of the lines which are at right angles.

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Solution

Equation of line :- $y + 2 a t = t x - a t^{3} \Rightarrow$ slope = t
Two perpendicular lines will have slopes
m and $\frac{- 1}{m}$ respectively
$\Rightarrow y + 2 a m = m x - a m^{3}$ and $y - \frac{2 a}{m} = \frac{- x}{m} + \frac{a}{m^{3}}$
$\Rightarrow a m^{3} + (2 a - x) m + y = 0$ $\Rightarrow m^{3} y - 2 a m^{2} = - x m^{2} + a$
$\Rightarrow y m^{3} + (x - 2 a) m^{2} - a = 0$
Both are cubic in m, hence, both should have
coefficient in same ratio.
$\Rightarrow \frac{a}{y} = \frac{0}{x - 2 a} = \frac{2 a - x}{0} = \frac{y}{- a}$
$\Rightarrow x - 2 a = 0 \Rightarrow x = 2 a$ and $y = a, - y = a$
$\Rightarrow y = 0$
$\Rightarrow$ Intersection point would be $(2 a, 0)$

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