Question

The straight line y = 3x − 3 intersects the parabola y^2 = 12x at the points P and
Q. Show that P and Q lie on opposite side of the x-axis and calculate the finite
area bounded by the chord and the arc PQ of the parabola.

Answer 1

$\text{If \( y = 3x - 3 \) and $y^2 = 12x$ intersect then we will have the following: } \\ (3x - 3)^2 = 12 x \implies 9x^2 -18x -9 = 12x \\ \implies 9x^2 - 30x - 9 = 0 \implies (x-3)(9x -3) = 0 \implies x = 3 \text{ or } x = \frac{1}{3} \\ \text{Putting these into \( y = 3x - 3\) we have the following: }\\ \text{When $x = 3$ \, \, $y = 3(3) - 3 = 6$} \\ \text{When $x = \frac 13$ \, \, $y = 3(\frac 13) - 3 = -2$}\\ \text{Then we have that point $P = (3,6)$ and point $Q = (\frac 13, -2)$} \\ \text{Since both the $x$ coordinates of $P$ and $Q$ are positive but their $y$ coordinates have values of different signs hence it is enough to say that $P$ and $Q$ lies on opposite side of $x$ axis} \\ \text{Note that from $y^2 = 12x$ we have that $y = \sqrt{12x}$} \\ \text{Area bounded = } \int_{\frac13}^{3} \left[ \sqrt{12x} - (3x - 3) \right] \,dx = \int_{\frac13}^{3}\left[ 2\sqrt{3x} - 3x + 3 \right]dx = \left[\frac{4\sqrt3}{3}x\sqrt x - \frac32 x^2 + 3x\right]_{\frac13}^{3} \\ \qquad \qquad \qquad \,\,= \left[\frac{4\sqrt3}{3}3\sqrt 3 - \frac32 (3)^2 + 3(3)\right] - \left[\frac{4\sqrt3}{3}(\frac13)\sqrt {\frac13} - \frac32 (\frac13)^2 + 3(\frac13)\right] = 12 - \frac{27}{2} + 9- \frac49 +\frac16 - 1 \\ \qquad \qquad \qquad \,\,= \frac{56}{9} \text{units square}$