A tangent is drawn to the parabola $y^{2}$ = 4x at the point 'p' whose abscissa lies in the interval $[1, 4]$ . The maximum possible area of the triangle formed by tangent at 'p', ordinate of the point 'p' and the x-axis is equal to?

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Solution

Consider the problem
$\begin{matrix} y^{2} = 4 x 4 a = 4 a = 1 \end{matrix}$

Let the point $P (t^{2}, 2 t)$

Given $1 \leq t^{2} \leq 4$

$\begin{matrix} t^{2} \geq 1 t \geq 1 o r t \leq - 1 t^{2} \leq 4 - 2 \leq t \leq 2....... (1) \end{matrix}$

And

$\begin{matrix} y^{2} = 4 x 2 y \frac{d y}{d x} = 4 m = \frac{4}{2 y} = \frac{2}{y} \end{matrix}$

equation of tangent at $P$

$y - 2 t = m (x - t^{2})$

Now put $y = 0$ for intersecting point with $X - a x i s$

$\begin{matrix} - 2 = \frac{2}{2 t} (x - t^{2}) - 2 t^{2} = x - t^{2} x = - t^{2} \end{matrix}$

$\begin{matrix} M = (- t^{2}, 0) B = (t^{2}, 0) B M = 2 t^{2} \end{matrix}$

So,Length of ordinate $A B = 2 t$

Since, $A B M$ is a right triangle
$A r e a o f t r i a n g l e = \frac{1}{2} \times b a s e \times h e i g h t$

So,
$\begin{matrix} A r e a = \frac{1}{2} \times 2 t \times 2 t^{2} = 2 t^{3} \end{matrix}$

And maximum value of $t$ from $(1)$ $= 2$
$\begin{matrix} A r e a = 2 \times {(2)}^{3} = 2 \times 8 = 16 s q . u n i t s \end{matrix}$

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