Let $y^2 = 4ax$ be a parabola and $PQ$ is a focal chord. Let $R$ be the point of intersection of the tangents at $P$ and $Q$, then:
area of circumcircle of $\triangle PQR$ is $\frac{\pi(PQ)^2}{4}$
orthocentre of $\triangle PQR$ lies at the directrix
incentre of $\triangle PQR$ lies at the vertex
minimum area of the circumcircle a $\triangle PQR$ is $4\pi a^2$
Step-by-Step Solution
Key Concept: For parabola y² = 4ax, tangents at endpoints P and Q of a focal chord intersect at point R on the directrix perpendicularly, forming a right angle. Since ∠PRQ = 90°, the focal chord PQ becomes the diameter of the circumcircle, giving circumradius = PQ/2 and circumcircle area = π(PQ)²/4.
For a parabola $y^2 = 4ax$, tangents at endpoints of a focal chord intersect at the directrix perpendicularly. The orthocenter of the right triangle formed by tangent intersection points lies on the directrix. The circumcircle of triangle $PQR$ has diameter $PQ$ (the focal chord). Minimum focal chord length equals the latus rectum $4a$, so minimum circumcircle area is $\pi(4a)^2 = 4\pi a^2$. For tangents $t_1y = x + at_1^2$ and $t_2y = x + at_2^2$ with $t_1t_2 = -1$, distances from origin are $d_1 = \frac{at_1^2}{\sqrt{1+t_1^2}}$ and $d_2 = \frac{at_2^2}{\sqrt{1+t_2^2}}$, which are unequal, so the incenter cannot lie at the vertex.
Correct Answer: 1,2,4