Question

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

• A. (4a, 0)
• B. (2a, 0)
• C. (- 4a, 0)
• D. ( - 2a, 0)

Answer 1

Answer: (A)

On homogenisation of the curve $y^2-4ax=0$ by line $y=mx+c$, we are getting combined equation of straight lines which subtend a right angle at the origin, So

$y^2-4ax\left(\frac{y-mx}{c}\right)=0$
$\Rightarrow c+4am=0$ ...(i)
On putting the value of ${}^{\prime }{c}^{\prime }$ in the line, we get $y=m(x-4a)$, represent family of line passes through $(4a,0)$.