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(C) Step $1$: Identify the point on the parabola.The parabola is $y^2 = 4ax$. The end of the latus rectum is $P(a, 2a)$.Step $2$: Find the equations o

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$4a^2$

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(C) Step $1$: Identify the point on the parabola.The parabola is $y^2 = 4ax$. The end of the latus rectum is $P(a, 2a)$.Step $2$: Find the equations of the tangent and normal.The slope of the tangent at $P(a, 2a)$ is given by $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2a} = 1$.The equation of the tangent at $P(a, 2a)$ is $y - 2a = 1(x - a) \Rightarrow y = x + a$.This tangent meets the $x$-axis $(y=0)$ at $T(-a, 0)$.The slope of the normal at $P(a, 2a)$ is $-1$ (since the tangent slope is $1$).The equation of the normal at $P(a, 2a)$ is $y - 2a = -1(x - a) \Rightarrow y = -x + 3a$.This normal meets the $x$-axis $(y=0)$ at $N(3a, 0)$.Step $3$: Calculate the area of the triangle $PTN$.The vertices of the triangle are $P(a, 2a)$,$T(-a, 0)$,and $N(3a, 0)$.The base $TN$ lies on the $x$-axis,with length $TN = |3a - (-a)| = 4a$.The height of the triangle is the $y$-coordinate of $P$,which is $2a$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4a 	imes 2a = 4a^2$.

Find the area of the triangle formed by the tangent and the normal at one end of the latus rectum of the parabola $y^2 = 4ax$ with the axis of the parabola.

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