JEE Main · 2018 · 08 Apr 2018 · Offline · Parabola

On $y^2=16x$ the slope at $P(16,16)$ is $\frac{dy}{dx}=\frac{8}{y}=\frac{1}{2}$, so the tangent meets the axis at $A(-16,0)$ and the normal, of slope $-2$, meets it at $B(24,0)$. Since the tangent and normal are perpendicular, the angle at $P$ in triangle $PAB$ is a right angle and $AB$ is the diameter of the circle through $P,A,B$, so its centre is the midpoint $C(4,0)$, the focus of the parabola. Then $m_{CP}=\frac{16-0}{16-4}=\frac{4}{3}$ and $m_{PB}=\frac{16-0}{16-24}=-2$, so $\tan \theta =\left|\frac{\frac{4}{3}-(-2)}{1+\frac{4}{3}(-2)}\right|=\left|\frac{10/3}{-5/3}\right|=2$.