 # Problem of the Week Problem E and Solution Parabolic Art

## Problem

Kenna likes making artistic creations using parabolas, to put on the walls of her math classroom. She drew a parabola with vertex $$E(7,9)$$ and plotted points $$A(9,8)$$ and $$B(3,b)$$ on the parabola as well as points $$C$$ and $$D$$ where the parabola intersects the $$x$$-axis, with $$C$$ to the left of $$D$$. Then she connected points $$A$$, $$B$$, $$C$$, and $$D$$ to form quadrilateral $$ABCD$$, and painted it blue. What is the area of quadrilateral $$ABCD$$? ## Solution

First we need to find the equation of the parabola. Then, we can find the $$x$$-intercepts of the parabola and the $$y$$-coordinate of point $$B$$ on the parabola.

We are given the vertex of the parabola, $$E(7,9)$$. Using the vertex form of the equation of a parabola, $$y=a(x-h)^2+k$$, with vertex $$(h,k)=(7,9)$$, the equation of the parabola is $$y=a(x-7)^2+9$$.

Since the point $$A(9,8)$$ is on the parabola, we can substitute $$(x,y)=(9,8)$$ into the equation $$y=a(x-7)^2+9$$ to find the value of $$a$$. \begin{aligned} 8&=a(9-7)^2+9\\ 8&=a(4)+9\\ -1&=4a\\ -\frac{1}{4}&=a\end{aligned} The equation of the parabola is therefore $$y=-\frac{1}{4}(x-7)^2+9$$.

To find the $$y$$-coordinate of $$B(3,b)$$, we substitute $$(x,y)=(3,b)$$ into the equation of the parabola. \begin{aligned} b&=-\frac{1}{4}(3-7)^2+9\\ &=-\frac{1}{4}(16)+9\\ &=-4+9\\ &=5\end{aligned}

Therefore, the coordinates of $$B$$ are $$(3,5)$$.

To find the $$x$$-intercepts of the parabola, we substitute $$y=0$$ into the equation of the parabola. \begin{aligned} 0&=-\frac{1}{4}(x-7)^2+9\\ -9&=-\frac{1}{4}(x-7)^2\\ 36&=(x-7)^2\\ \pm 6&=x-7\end{aligned} It follows that $$x-7=-6$$ or $$x-7=6$$. Then the $$x$$-intercepts of the parabola are 1 and 13. Therefore, the coordinates of $$C$$ and $$D$$ are $$C(1,0)$$ and $$D(13,0)$$.

Now that we know the coordinates of $$A$$, $$B$$, $$C$$, and $$D$$, we can calculate the area of quadrilateral $$ABCD$$. There are many ways to do this. We will proceed as follows.

From $$B(3,5)$$ and $$A(9,8)$$, drop perpendiculars, intersecting the $$x$$-axis at $$F(3,0)$$ and $$G(9,0)$$, respectively. From $$B(3,5)$$ draw a line perpendicular to $$AG$$, intersecting $$AG$$ at $$H(9,5)$$. Draw line segment $$BG$$. Note that line segments $$BG$$ and $$AG$$ divide the quadrilateral into three regions: $$\triangle CGB$$, $$\triangle AGD$$, and $$\triangle AGB$$.

We will use the coordinates of the points to find the lengths of several horizontal and vertical line segments that will be required for the area calculation. $BH=9-3=6,\ CG=9-1=8,\ GD=13-9=4,\ BF=5-0=5,\text{ and }AG=8-0=8.$

To determine the area of $$ABCD$$, we will find the sum of the areas of $$\triangle CGB$$, $$\triangle AGD$$ and $$\triangle AGB$$. \begin{align*} \text{Area }ABCD&=\text{Area }\triangle CGB+\text{Area }\triangle AGD+\text{Area }\triangle AGB\\ &=\frac{CG\times BF}{2}+\frac{AG\times GD}{2}+\frac{AG\times BH}{2}\\ &=\frac{8\times 5}{2}+\frac{8\times 4}{2}+\frac{8\times 6}{2}\\ &=20+16+24\\ &=60\text{ units}^2 \end{align*} Therefore, the area of quadrilateral $$ABCD$$ is $$60\text{ units}^2.$$