Question

If the parabola $y^2=4x$ meets a circle with centre at $(6,5)$ orthogonally, then possible point (s) of intersection can be;

• A. $(4,4)$
• B. $(9,4)$
• C. $(2,8)$
• D. $(3,2)$

Answer 1

Answer: (A)

$(4,4)$

Let the point of intersection be $(t^2,2t)$

$y^2=4x$

$\therefore \frac{dy}{dx}=\frac{2}{y}$

Slope of tangent to parabola at $(t^2,2t)=\frac{2}{2t}=\frac{1}{t}$

Since the circle and parabola meet orthogonally, tangent to circle and parabola at $(t^2,2t)$ are perpendicular.

Let $m$ be slope of tangent to circle at $(t^2,2t)$

$\therefore m\frac{1}{t}=-1$

$\therefore m=-t$

So, slope of normal to circle $(t^2,2t)=\frac{-1}{-t}=\frac{1}{t}$

$\therefore \frac{2t-5}{t^2-6}=\frac{1}{t}$

$\therefore t^2-5t+6=0$

$⟹t=2$ and $t=3$

So, the possible points of intersection are $(4,4)$ and $(9,6)$

So, the answer is option (A).