Question

TP and TQ are tangents to parabola $y^2 = 8x$ and normals at $P$ and $Q$ intersect at a point $R$ on the parabola. The locus of circumcentre of $\Delta TPQ$ is a parabola whose:

• A. Vertex is $(2, 0)$
• B. Foot of perpendicular from focus on the directrix is $\left(\tfrac{7}{4}, 0\right)$
• C. Length of latus rectum is 1
• D. Focus is $\left(\tfrac{9}{4}, 0\right)$

Answer 1

Answer: (A), (B), (C), (D)

Vertex is $(2, 0)$; Foot of perpendicular from focus on the directrix is $\left(\tfrac{7}{4}, 0\right)$; Length of latus rectum is 1; Focus is $\left(\tfrac{9}{4}, 0\right)$

Answer 2

Solution Outline

1. Given parabola: $y^2=8x$ has parameter $a=2$.
2. Construction: Let the external point be $T$. Tangents from $T$ touch at $P$ and $Q$. Normals at $P,Q$ meet on the parabola, ensuring a symmetry that makes $T$, $P$, $Q$ equidistant from the circumcentre.
3. Locus derivation (sketch):
   • One finds that the circumcentre $(X,Y)$ satisfies $$Y^2 \;=\; X-2,$$ i.e.\ the new parabola is $$y^2 \;=\;(x-2).$$
4. Extracting key parameters of $y^2 = (x-2)$:
   • Vertex $(h,0) = (2,0)$.
   • Standard form $y^2 = 4a\,(x-h)$ gives $4a=1\implies a=\tfrac14$.
   • Focus $\bigl(h+a,0\bigr) = \bigl(2+\tfrac14,0\bigr) = \bigl(\tfrac94,0\bigr)$.
   • Directrix $x = h - a = 2 - \tfrac14 = \tfrac74$.
   • Foot of perpendicular from focus $\bigl(\tfrac94,0\bigr)$ onto directrix $x = \tfrac74$ is $\bigl(\tfrac74,0\bigr)$.
   • Latus rectum length $=4a=1$.

All four statements (A), (B), (C) and (D) are thus correct.