Let P(4, - 4) and Q(9, 6) be two points on the parabola, y ^ 2 = 4x...

Two different approaches we can use here.

Approach 1:

Let $X$ be \left({t}^{2}, 2t\right)$, then</p><p>Area of$\delta PXQ = \dfrac{1}{2}\left|$\begin{cases}{t}^{2} & 2t & 1 \\ 9 & 6 & 1 \\ 4 & - 4 & 1\end{cases}$\right|$</p><p>$\Delta = \dfrac{1}{2} . 10\left|{t}^{2} - t - 6\right| . . . \left(i\right)$</p><p>For maxima, differentiating both the sides with respect to$t$and equating it to zero, we get</p><p>${\Delta}^{′} = 0 ⇒ 5\left(2t - 1\right) = 0 ⇒ t = \dfrac{1}{2}$</p><p>Hence, area of$\delta PXQ = 5\left|{\left(\dfrac{1}{2}\right)}^{2} - \dfrac{1}{2} - 6\right| = 5\left|\dfrac{1}{4} - \dfrac{1}{2} - 6\right| = 5\left|\dfrac{1 - 2 - 24}{4}\right| = \dfrac{125}{4}$sq.units (using equation \left(i\right)$)

Approach 2:

For maximum area tangent to the parabola at $X$ must be parallel to $PQ$. Let $X\left({t}^{2}, 2t\right)$, then

$2y\dfrac{dy}{dx} = 4 ⇒ \dfrac{dy}{dx} = \dfrac{2}{y} ⇒ {\left(\dfrac{dy}{dx}\right)}_{\left({t}^{2}, 2t\right)} = \dfrac{1}{t}$

Also, slope of line $PQ = \dfrac{6 + 4}{9 - 4} = 2$

Since, both the slopes are equal.

Thus, $t = \dfrac{1}{2} ⇒ X\left(\dfrac{1}{4}, 1\right)$

Therefore, area of \delta PXQ = \dfrac{1}{2}\left|\begin{cases}4 & - 4 & 1 \\ \dfrac{1}{4} & 1 & 1 \\ 9 & 6 & 1\end{cases}\right| = \dfrac{1}{2}\left|4\left(1 - 6\right) - \left(- 4\right)\left(\dfrac{1}{4} - 9\right) + 1\left(\dfrac{3}{2} - 9\right)\right| = \dfrac{1}{2}\left|- 20 - 35 - \dfrac{15}{2}\right| = \dfrac{125}{4} sq.units.