Parabola
Three Points on Parabola — Perpendicular Feet Ratio
DAILY_CHALLENGE
Grade 11
Question:
Let $A$, $B$ and $C$ be three points on the parabola $y^2=6x$ and let the line segment $AB$ meet the line $L$ through $C$ parallel to the $x$-axis at the point $D$. Let $M$ and $N$ respectively be the feet of the perpendiculars from $A$ and $B$ on $L$. Then $\left(\dfrac{AM\cdot BN}{CD}\right)^2$ is equal to
Step-by-Step Solution
Key Concept: Parametrize: $A=(at_1^2,2at_1)$, $B=(at_2^2,2at_2)$, $C=(at_3^2,2at_3)$ on $y^2=6x$ ($a=3/2$). $L$: $y=2at_3$. $D$ is intersection of line $AB$ with $L$. $AM=|2a(t_1-t_3)|$, $BN=|2a(t_2-t_3)|$, $CD=|at_3^2-\alpha|$ where $\alpha$ is the $x$-coordinate of $D$.
$\left(\frac{AM\cdot BN}{CD}\right)^2=16a^2=36$.
Correct Answer: 36