The equations of the normals at the ends of the latus rectum of the parabola y² = 4ax are given by

The correct answer is: x² – y² – 6ax + 9a² = 0

The co-ordinates of end points of latus rectum is (a, 2a) and (a,-2a) .
Then slope of tangent of y2 = 4ax is  
Slope of tangent at (a, 2a) and (a,-2a) are 1 and -1 respectively.
As, Product of Slopes of perpendicular lines is -1.
Then Slope of Normal at (a, 2a) and (a,-2a) are  and   respectively.
Equations of normal are y - 2a = -1(x-a) => y+ x -3a = 0 and y + 2a = 1(x-a) => x-y -3a =0
Then equation of pair of lines is  (y+ x -3a)( x-y -3a ) = 0
The equations of the normals at the ends of the latus rectum of the parabola y2 = 4ax are given by
x² - y²- 6ax + 9a² = 0 .