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In the image below, $P$ and $Q$ each have a $y$ coordinate of $3$ .

Triangle drawn in parabola

The parabola has equation

y + 9 = x^{2} + 4 x

The area of $△ P Q R$ is $A$ , and the perimeter can be given in the form

p (1 + \sqrt{q})

where $q$ is prime.

Find the value of

A p q

Hint

To find $P$ and $Q$ , set $y = 3$ and solve for $x$ .

Hint

To find the perimeter, you will need to use Pythagoras' Theorem

Solution

At $P$ and $Q$ we have $y = 3$ , so

\begin{aligned} 12 & = x^{2} + 4 x \\ x^{2} + 4 x - 12 & = 0 \\ (x + 6) (x - 2) & = 0 \\ x & \in {- 6, 2} \end{aligned}

and that means $P (- 6, 3)$ and $Q (2, 3)$ .

At $R$ , we have $x = \frac{- 6 + 2}{2} = - 2$ and so $y = x^{2} + 4 x - 9 = - 13$ . Thus $R (- 2, - 13)$ .

Parabola with triangle

The height of the triangle is $3 - (- 13) = 16$ and the base is $2 - (- 6) = 8$ .

The area is given by $\frac{16 \times 8}{2} = 64$ .

For the perimeter, we need the length of the other two sides of the triangle.

By Pythagoras, this is

\sqrt{4^{2} + 16^{2}} = 4 \sqrt{17}

and so the perimeter is

8 + 4 \sqrt{17} + 4 \sqrt{17} = 8 (1 + \sqrt{17})