The area of the triangle formed by the tangent and the normal to the parabola y2=4ax, both drawn at the same end of the latus rectum, and the axis of the parabola is
4a2
One end of the latus rectum is P(a, 2a)
The equation of the tangent PT at P(a,2a) is
y.2a = 2a(x+a), i.e., y =x+a.
The equation of normal PN at (a, 2a) is y + x =2a+a, i.e., y + x =3a.
Solving y=0 and y=x+a, we get x=-a, y=0.
Solving y=0, y+x=3a, we get x=3a, y=0.
The area of the triangle with vertices P(a, 2a), T(-a, 0), N(3a, 0) is 4a2.