Let P(4,-4) and Q(9,6) be two points on the parabola, y^2=4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this…

Two different approaches we can use here.

Approach 1:

Let $X$X be $({t}^{2},2t)$(t2,2t), then

Area of $\Delta PXQ=\frac{1}{2}|\begin{matrix}{t}^{2} & 2t & 1 \\ 9 & 6 & 1 \\ 4 & -4 & 1\end{matrix}|$ΔPXQ=21​∣t294​2t6−4​111​∣

$$\Delta =\frac{1}{2}.10|{t}^{2}-t-6|...(i)$$Δ=21​.10∣t2−t−6∣...(i)

For maxima, differentiating both the sides with respect to $t$t and equating it to zero, we get

$${\Delta }^{'}=0\Rightarrow 5(2t-1)=0\Rightarrow t=\frac{1}{2}$$Δ′=0⇒5(2t−1)=0⇒t=21​

Hence, area of $\Delta PXQ=5|{(\frac{1}{2})}^{2}-\frac{1}{2}-6|=5|\frac{1}{4}-\frac{1}{2}-6|=5|\frac{1-2-24}{4}|=\frac{125}{4}$ΔPXQ=5∣(21​)2−21​−6∣=5∣41​−21​−6∣=5∣41−2−24​∣=4125​ sq.units (using equation $(i)$(i))

Approach 2:

For maximum area tangent to the parabola at $X$X must be parallel to $PQ$PQ. Let $X({t}^{2},2t)$X(t2,2t), then

$$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}\Rightarrow {(\frac{dy}{dx})}_{({t}^{2},2t)}=\frac{1}{t}$$2ydxdy​=4⇒dxdy​=y2​⇒(dxdy​)(t2,2t)​=t1​

Also, slope of line $PQ=\frac{6+4}{9-4}=2$PQ=9−46+4​=2

Since, both the slopes are equal.

Thus, $t=\frac{1}{2}\Rightarrow X(\frac{1}{4},1)$t=21​⇒X(41​,1)

Therefore, area of $\Delta PXQ=\frac{1}{2}|\begin{matrix}4 & -4 & 1 \\ \frac{1}{4} & 1 & 1 \\ 9 & 6 & 1\end{matrix}|=\frac{1}{2}|4(1-6)-(-4)(\frac{1}{4}-9)+1(\frac{3}{2}-9)|=\frac{1}{2}|-20-35-\frac{15}{2}|=\frac{125}{4}$ΔPXQ=21​∣441​9​−416​111​∣=21​∣4(1−6)−(−4)(41​−9)+1(23​−9)∣=21​∣−20−35−215​∣=4125​ sq.units.