Subfield: Algebra


Question: Suppose that $a>\frac{1}{2}$ and that the parabola with equation $y=a x^{2}+2$ has vertex $V$. The parabola intersects the line with equation $y=-x+4 a$ at points $B$ and $C$, as shown. If the area of $\triangle V B C$ is $\frac{72}{5}$, determine the value of $a$.

<img_3292>

Solution:
The parabola with equation $y=a x^{2}+2$ is symmetric about the $y$-axis.

Thus, its vertex occurs when $x=0$ (which gives $y=a \cdot 0^{2}+2=2$ ) and so $V$ has coordinates $(0,2)$.

To find the coordinates of $B$ and $C$, we use the equations of the parabola and line to obtain

$$
\begin{aligned}
a x^{2}+2 & =-x+4 a \\
a x^{2}+x+(2-4 a) & =0
\end{aligned}
$$

Using the quadratic formula,

$$
x=\frac{-1 \pm \sqrt{1^{2}-4 a(2-4 a)}}{2 a}=\frac{-1 \pm \sqrt{1-8 a+16 a^{2}}}{2 a}
$$

Since $1-8 a+16 a^{2}=(4 a-1)^{2}$ and $4 a-1>0\left(\right.$ since $\left.a>\frac{1}{2}\right)$, then $\sqrt{1-8 a+16 a^{2}}=4 a-1$ and so

$$
x=\frac{-1 \pm(4 a-1)}{2 a}
$$

which means that $x=\frac{4 a-2}{2 a}=\frac{2 a-1}{a}=2-\frac{1}{a}$ or $x=\frac{-4 a}{2 a}=-2$.

We can use the equation of the line to obtain the $y$-coordinates of $B$ and $C$.

When $x=-2$ (corresponding to point $B$ ), we obtain $y=-(-2)+4 a=4 a+2$.

When $x=2-\frac{1}{a}$ (corresponding to point $C$ ), we obtain $y=-\left(2-\frac{1}{a}\right)+4 a=4 a-2+\frac{1}{a}$.

Let $P$ and $Q$ be the points on the horizontal line through $V$ so that $B P$ and $C Q$ are perpendicular to $P Q$.

<img_3382>



Then the area of $\triangle V B C$ is equal to the area of trapezoid $P B C Q$ minus the areas of right-angled $\triangle B P V$ and right-angled $\triangle C Q V$.

Since $B$ has coordinates $(-2,4 a+2), P$ has coordinates $(-2,2), V$ has coordiantes $(0,2)$, $Q$ has coordinates $\left(2-\frac{1}{a}, 2\right)$, and $C$ has coordinates $\left(2-\frac{1}{a}, 4 a-2+\frac{1}{a}\right)$, then

$$
\begin{aligned}
B P & =(4 a+2)-2=4 a \\
C Q & =\left(4 a-2+\frac{1}{a}\right)-2=4 a-4+\frac{1}{a} \\
P V & =0-(-2)=2 \\
Q V & =2-\frac{1}{a}-0=2-\frac{1}{a} \\
P Q & =P V+Q V=2+2-\frac{1}{a}=4-\frac{1}{a}
\end{aligned}
$$

Therefore, the area of trapezoid $P B C Q$ is

$$
\frac{1}{2}(B P+C Q)(P Q)=\frac{1}{2}\left(4 a+4 a-4+\frac{1}{a}\right)\left(4-\frac{1}{a}\right)=\left(4 a-2+\frac{1}{2 a}\right)\left(4-\frac{1}{a}\right)
$$

Also, the area of $\triangle B P V$ is $\frac{1}{2} \cdot B P \cdot P V=\frac{1}{2}(4 a)(2)=4 a$.

Furthermore, the area of $\triangle C Q V$ is

$$
\frac{1}{2} \cdot C Q \cdot Q V=\frac{1}{2}\left(4 a-4+\frac{1}{a}\right)\left(2-\frac{1}{a}\right)=\left(2 a-2+\frac{1}{2 a}\right)\left(2-\frac{1}{a}\right)
$$

From the given information,

$$
\left(4 a-2+\frac{1}{2 a}\right)\left(4-\frac{1}{a}\right)-4 a-\left(2 a-2+\frac{1}{2 a}\right)\left(2-\frac{1}{a}\right)=\frac{72}{5}
$$

Multiplying both sides by $2 a^{2}$, which we distribute through the factors on the left side as $2 a \cdot a$, we obtain

$$
\left(8 a^{2}-4 a+1\right)(4 a-1)-8 a^{3}-\left(4 a^{2}-4 a+1\right)(2 a-1)=\frac{144}{5} a^{2}
$$

Multiplying both sides by 5 , we obtain

$$
5\left(8 a^{2}-4 a+1\right)(4 a-1)-40 a^{3}-5\left(4 a^{2}-4 a+1\right)(2 a-1)=144 a^{2}
$$

Expanding and simplifying, we obtain

$$
\begin{aligned}
\left(160 a^{3}-120 a^{2}+40 a-5\right)-40 a^{3}-\left(40 a^{3}-60 a^{2}+30 a-5\right) & =144 a^{2} \\
80 a^{3}-204 a^{2}+10 a & =0 \\
2 a\left(40 a^{2}-102 a+5\right) & =0 \\
2 a(20 a-1)(2 a-5) & =0
\end{aligned}
$$

and so $a=0$ or $a=\frac{1}{20}$ or $a=\frac{5}{2}$. Since $a>\frac{1}{2}$, then $a=\frac{5}{2}$.

Final answer: $\frac{5}{2}$