Coordinate Geometry
Normal to parabola; area of triangle condition
MMTS_Full_Test_09
Grade 12
Question:
A normal with slope $\frac{1}{m}$ is drawn from $P(0,-k)$ to $x^2=-12y$. The line through $Q\!\left(0,-\frac{k}{506}\right)$ parallel to the tangent at vertex meets the parabola at $R,S$. Area of $\triangle ORS = 144$ sq. units. Then $m^2$ equals
(A) 2022
(B) 2024
(C) 2025
(D) 2026
Step-by-Step Solution
Key Concept: Normal to $x^2=-12y$: equation $x=my+6m+3m^3$ (using standard form). Condition $P(0,-k)$ gives $k=3(m^2+2)$. Area of $\triangle ORS$ using $Q$ and chord $RS$.
$m^2=2022$.
Correct Answer: (A) 2022