Parabola
Circumcentre locus and conic eccentricity
MJAT_TS1_P1
Grade 12
Question:
Let $A$, $B$ be variable points on the parabola $y^2 = 12x$, and let $N$, $K$ be the feet of perpendiculars from $A$, $B$ to the directrix respectively. Let $S$ be the focus of the parabola. Find the locus of the circumcentre of $\triangle NSK$ as $A$, $B$ vary on the parabola such that $\angle NSK = 60°$. If this locus is a conic of eccentricity $e$, then $(e-2)^2$ is:
Step-by-Step Solution
Key Concept: For any point $A$ on the parabola, $AS = AN$ (definition of parabola). So $\triangle ANS$ is isosceles: the tangent at $A$ is the perpendicular bisector of $NS$, meaning the circumcentre $C$ of $\triangle NSK$ lies on the tangent at $A$ (and similarly for $B$). Use this to find the locus.
$AS = AN \Rightarrow$ tangent at $A$ bisects $NS$ perpendicularly $\Rightarrow$ circumcentre $C$ satisfies $CK = CD$ and $CS/CD = 2$ (from the $60°$ condition). Locus is a hyperbola with $e = 2$. Thus $(e-2)^2 = 0$.
Correct Answer: 0