Find the area-No Parabolem!

Given that area of $\triangle ABC = 5$ .

Taking the base as $AC$ and height $OB$ ( $O$ is the origin), we have,

$\begin{aligned} &\dfrac12 \cdot (5-r) \cdot 5 = 5 \\ \implies & 5-r = 2 \\ \implies & r = 3 \end{aligned}$

So the point $A \equiv \left( 3,0 \right)$ .

We can observe that the axis of this parabola is parallel to the $y-axis$ and hence the abscissa of its vertex will be exactly in between $x=3$ and $x=5$ , thus giving $p=4$ .

There are three points given to us on the parabola. Using the equation $y = A x^2 + B x + C$ for the equation of this parabola and plugging in the corresponding values of $x$ and $y$ from the three given points $A,B$ and $C$ , we get,

$A=\dfrac13 ~ , ~ B=-\dfrac{8}{3} ~ , ~ C = 5$

So, our equation of parabola becomes,

$y = \dfrac{x^2}{3} - \dfrac{8x}{3} + 5$

Plugging in for $x=4$ we get $y=-\dfrac{1}{3}$ .

Thus, the coordinates of point $D(p,q)$ are $\left( 4, -\dfrac{1}{3} \right)$ .

Hence, the area of $\triangle DBC$ is,

$\begin{aligned} \Delta & = & \dfrac 12 \begin{vmatrix} 0 & 5 & 1 \\ 4 & {-1}/{3} & 1 \\ 5 & 0 & 1 \end{vmatrix} \\ & = & \dfrac{10}{3} = \boxed{3.33} \end{aligned}$