Coordinate Geometry
Parabola / Tangent
MMTS_Full_Test_01
Grade 12
Question:
A line $y = m(x-4)$ meets the $x$-axis at $P$ and the parabola $x^2 = 32y$ at $Q(x_1,y_1)$. The tangent to the parabola at $Q$ meets the $x$-axis at $R(x_2,0)$, $0 < x_2 < 6$. If the area of $\triangle PQR$ assumes a local maximum, then the value of $m$ is
$\frac{1}{3}$
$\frac{2}{3}$
$\frac{1}{2}$
$\frac{3}{4}$
Step-by-Step Solution
Key Concept: Point on $x^2=32y$: $(t, t^2/32)$; tangent at $t$: $xt=16y+t^2/2$... Express area as function of $t$ and maximize.
Line meets $x$-axis at $P=(4,0)$. On parabola: $x_1^2=32y_1$ and $y_1=m(x_1-4)$. Tangent at $Q$: $xx_1=16y+x_1^2/2$; meets $x$-axis: $x_2=x_1/2$. Area $\triangle PQR = \frac{1}{2}|PR||y_1|=\frac{1}{2}|4-x_1/2|\cdot y_1$. Maximizing over $x_1$ with constraint and $0<x_2=x_1/2<6$: $x_1\in(0,12)$. Max area gives $m=2/3$.
Correct Answer: B