Find the equations of normal to the parabola `y^2=4a x` at the ends of the latus rectum.

Options

A. `x^(2)-y^(2)-3ax+9s^(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

A

Explanation

The coordinates of the ends of the laturrectum 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=(-2a)/(2a)(x-a)`
or `x+y-3a=0`
Similarly, the equation of the normal (a, -2a) is
`x-y-3a=0`
The combained equation is
`x^(2)-y^(2)-6ax+9a^(2)=0`