Question

The equations of the normals at the ends of the latusrectum of the parabola $ y = 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

Answer: (A)

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