Variable straight lines $y=m x+c$ make intercepts on the curve $y^2-4 a x=0$ which subtend a right angle at the origin. Then the point of concurrence of these lines $y=m x+c$ is

Question

Variable straight lines $y=m x+c$ make intercepts on the curve $y^2-4 a x=0$ which subtend a right angle at the origin. Then the point of concurrence of these lines $y=m x+c$ is, $(4 a, 0)$, $(2 a, 0)$, $(-4 a, 0)$, $(-2 a, 0)$

MARKS App · Accepted Answer

On homogenisation of the curve $y^2-4 a x=0$ by line $y=m x+c$, we are getting combined equation of straight lines which subtend a right angle at the origin, so $\begin{aligned} \Rightarrow \quad y^2-4 a x\left(\frac{y-m x}{c}\right) & =0 \ c+4 a m & =0 \quad \ldots (i) \end{aligned}$ On putting the value of ' $c$ ' in the line, we get $y=m(x-4 a)$, represent family of line passes through $(4 a, 0)$. Hence, option (1) is correct.

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