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

• Correct answer
  A

  \((4 a, 0)\)
• B

  \((2 a, 0)\)
• C

  \((-4 a, 0)\)
• D

  \((-2 a, 0)\)

Solution

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.

Hence, the correct answer is option: 'A'.