Coordinate Geometry
Concurrent normals; centroid of medial triangle
MMTS_Full_Test_19
Grade 12
Question:
Let $A$, $B$, $C$ be 3 points on the parabola $y^2=4x$. Let $D$, $E$, $F$ be midpoints of $AB$, $BC$ and $AC$ respectively. If $G$ is the centroid of $\triangle DEF$ and normals to the parabola at $A$, $B$ and $C$ are concurrent at $H(5,1)$, then slope of the line $GH$ is
(A) $\dfrac{1}{2}$
(B) $\dfrac{1}{3}$
(C) $1$
(D) $\dfrac{3}{4}$
Step-by-Step Solution
Key Concept: The centroid of $\triangle ABC$ equals the centroid of $\triangle DEF$ (medial triangle). For normals concurrent at $(h,k)$: sum of parameters $t_1+t_2+t_3=0$, centroid of $\triangle ABC = (h-2, 0)=(3,0)$... actually $G=(h-2,0)=(3,0)$. Wait: centroid $\bar x = (t_1^2+t_2^2+t_3^2)/3$. Use $\sum t_i=0$ and other relations.
$G=(2,0)$, $H=(5,1)$. Slope $=1/3$.
Correct Answer: (B) $\dfrac{1}{3}$