Question

The equations of the normals at the ends of the latus-rectum of the parabola $y^2=4ax$ are given by

• A. $x^2-y^2-6ax+9a^2=0$
• B. $x^2-y^2-6ax-6ay+9a^2=0$
• C. $x^2-y^2-6ay+9a^2=0$
• D. None of the above

Answer 1

The correct option is A $x^2-y^2-6ax+9a^2=0$

The coordinates of the ends of the latus-rectum of the parabola $y^2=4ax$ are $(a,2a)$ respectively. The equation of the normal at $(a,2a)$ to $y^2=4ax$ is
$y-2a=\frac{-2a}{2a}(x-a)$
$x+y-3a=0$
Similarly, the equation of the normal $(a,-2a)$ is
$x-y-3a=0$
Then combined equation is
$x^2-y^2-6ax+9a^2=0$