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Q. The area of the triangle formed by the tangent and the normal to the parabola $y^{2}=4 a x$, both drawn at the same end of the latus rectum, and the axis of the parabola is

Conic Sections

Solution:

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One end of the latus rectum is $P(a, 2 a)$.
The equation of tangent $P T$ at $P(a, 2 a)$ is
$2 y a=2 a(x+a) $ or $ y=x+a$
The equation of normal $P N$ at $(a, 2 a)$ is
$y+x=2 a+a $ or $ y+x=3 a$
Solving $y=0$ and $y=x+a$, we get
$x=a, y=0$
Solving $y=0$ and $y+x=3 a$, we get
$x=3 a, y=0$
The area of the triangle with vertices $P(a, 2 a), T(a, 0)$, and $N(3 a, 0)$ is $4 a^{2}$.