A line with equation y = -3x + 9 intersects the axes at the points P and Q. A parabola of the form…
A line with equation $y = -3x + 9$ intersects the axes at the points P and Q. A parabola of the form $y = ax^2 + c$, where $a, c \in \mathbb{Z}$, also passes through the points P and Q as shown in the diagram below.
Parabola and Line
(a)[1]
Obtain the equation of the parabola.
(b)[2]
Using integration find the area of the shaded region.
Answer
Answer (a)
Official answer key
Putting $x = 0$ in $y = -3x + 9$ gives $P(0, 9)$; putting $y = 0$ gives $Q(3, 0)$.
The parabola $y = ax^2 + c$ passes through $(0, 9)$, so $c = 9$; through $(3, 0)$: $9a + 9 = 0$, so $a = -1$.
Equation of the parabola: $y = -x^2 + 9$.
Answer (b)
Official answer key
The shaded region lies between the parabola (above) and the line $y = 9 - 3x$ (below) from $x = 0$ to $x = 3$.
Area $= \int_0^3 \left[(9 - x^2) - (9 - 3x)\right]dx = \int_0^3 (3x - x^2)\,dx = \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{27}{2} - 9 = \frac{9}{2}$.
