Parabola — Normal and Area
DAILY_CHALLENGE
Grade None
Question:
Let $P$ be a point on the parabola $y^2=4ax$, where $a>0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $PFQ$, where $F$ is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a,m)$ is
$(2,3)$
$(1,3)$
$(2,4)$
$(3,4)$
Step-by-Step Solution
Key Concept: Area of triangle PFQ using base FQ along x-axis and height = y-coordinate of P
Parametrize: $P=(at^2,2at)$. Normal at $P$: $y=-tx+2at+at^3$, meets $x$-axis at $Q=(2a+at^2,0)$.
$F=(a,0)$. $FQ=|2a+at^2-a|=a(1+t^2)$. Height from $P$ to $x$-axis $=|2at|=2at$ (for $t>0$).
Area $=\dfrac{1}{2}\cdot a(1+t^2)\cdot2at=a^2t(1+t^2)=120$.
Slope of normal $=m=-t$ (for positive $m$, take $t=-m$):
$a^2m(1+m^2)=120$.
Test positive integer pairs:
$m=1$: $2a^2=120\Rightarrow a^2=60$. Not integer.
$m=2$: $10a^2=120\Rightarrow a^2=12$. Not integer.
$m=3$: $30a^2=120\Rightarrow a^2=4\Rightarrow a=2$. ✓
Answer: $(a,m)=(2,3)$.
Correct Answer: A