Question

Let $y^2-4ax$ be a parabola and PQ be a focal chord of parabola. Let T be the point of intersection of tangents at P and Q. Then.

• A. Area of circumcircle of $\Delta$PQT is $\frac{\pi P{Q}^2}{4}$
• B. Orthocentre of $\Delta$PQT will lie on tangent at vertex
• C. Incentre of $\Delta$PQT will be vertex of parabola
• D. Incentre of $\Delta$PQT will lie on directrix of parabola

Answer 1

Answer: (A)

Area of circumcircle of $\Delta$PQT is $\frac{\pi P{Q}^2}{4}$

$P$ has coordinates $(a{t}^2,2at)$ and $Q$ has coordinates $(a{t}_1^2,2a{t}_1)$ such that $t{t}_1=-1$.

Tangent at $P$ is $ty=x+a{t}^2$

Tangent at $Q$ is $t_{1}y=x+a{t}_1^2⟹-\frac{1}{t}y=x+\frac{a}{t^2}$

The meeting point of these tangents is $T(a,a(t+\frac{1}{t}))$

Also, these tangents are perpendicular to each other as the multiplication of their slopes is $-1$.

Hence, the circumcircle of right-angled $△PQT$ is circle passing through $P,Q,T$ and has centre at mid-point of $PQ$. Hence, area of this circle is $\frac{\pi (PQ{)}^2}{4}$.

Orthocenter of $△PQT$ is $T$ which does not lie on tangent at vertex.

Incentre lies in the triangle itself. But, the triangle lies completely to the right of tangent at vertex of parabola. Hence, the other 2 options are wrong.