Tangent and normal are drawn at P(16, 16) on the parabola $$y^2 = 16x$$, which intersect the axis of the parabola at A & B, respectively. If C is the center of the circle through the points P, A & B and $$\angle CPB = \theta$$, then a value of $$\tan \theta$$ is:

The parabola is $$y^{2}=16x$$. Comparing with the standard form $$y^{2}=4ax$$ we have $$4a=16\; \Rightarrow\; a=4$$. Its axis is the $$x$$-axis, i.e. the line $$y=0$$.

The given point is $$P(16,16)$$. First we confirm that it lies on the parabola:

$$y^{2}=16^{2}=256,\qquad 16x=16\cdot16=256;\qquad\text{hence }P\text{ is on the curve.}$$

We now require the equations of the tangent and the normal at $$P$$.

For $$y^{2}=16x$$ we differentiate implicitly:

$$2y\,\dfrac{dy}{dx}=16\quad\Longrightarrow\quad\dfrac{dy}{dx}=\dfrac{8}{y}.$$

At $$P(16,16)$$ the slope of the tangent is

$$m_{t}=\dfrac{8}{16}=\dfrac12.$$

The tangent therefore has the equation

$$y-16=\dfrac12\,(x-16).$$

To find its intersection with the axis $$y=0$$ we put $$y=0$$:

$$0-16=\dfrac12\,(x-16)\;\Longrightarrow\;-16=\dfrac{x-16}{2}$$

$$\Longrightarrow\;-32=x-16\;\Longrightarrow\;x=-16.$$

Thus the tangent meets the axis at $$A(-16,0).$$

The slope of the normal, being the negative reciprocal of the tangent’s slope, is

$$m_{n}=-\dfrac{1}{m_{t}}=-\dfrac{1}{\tfrac12}=-2.$$

The normal passes through $$P(16,16)$$, so

$$y-16=-2\,(x-16).$$

Setting $$y=0$$ to locate its intersection with the axis:

$$0-16=-2\,(x-16)\;\Longrightarrow\;-16=-2x+32$$

$$\Longrightarrow\;-48=-2x\;\Longrightarrow\;x=24.$$

Hence the normal meets the axis at $$B(24,0).$$

We now have the three points

$$A(-16,0),\; B(24,0),\; P(16,16).$$

Let $$C(h,k)$$ be the centre of the circle through $$A,B,P$$. Because $$A$$ and $$B$$ lie on the $$x$$-axis and are symmetric with respect to the centre, the perpendicular bisector of $$AB$$ is vertical.

The midpoint of $$AB$$ is

$$M_{1}\left(\dfrac{-16+24}{2},\dfrac{0+0}{2}\right)=(4,0).$$

Since $$AB$$ is horizontal, its perpendicular bisector is the vertical line $$x=4$$, so $$h=4$$. Thus $$C=(4,k).$$

Next we take the perpendicular bisector of $$BP$$ to determine $$k$$. The midpoint of $$BP$$ is

$$M_{2}\left(\dfrac{24+16}{2},\dfrac{0+16}{2}\right)=(20,8).$$

The slope of $$BP$$ is

$$m_{BP}=\dfrac{0-16}{24-16}=-\dfrac{16}{8}=-2,$$

so the slope of its perpendicular bisector is $$\dfrac12$$.

The equation of this bisector is therefore

$$y-8=\dfrac12\,(x-20).$$

Substituting $$x=4$$ into this equation (because $$C$$ lies on both bisectors) gives

$$y-8=\dfrac12\,(4-20)=\dfrac12(-16)=-8 \;\Longrightarrow\;y=0.$$

Hence $$k=0$$ and the centre is

$$C(4,0).$$

We now compute the angle $$\theta=\angle CPB$$, i.e. the angle at $$P$$ formed by the segments $$PC$$ and $$PB$$. For this we form their direction vectors:

$$\overrightarrow{PC}=C-P=(4-16,\,0-16)=(-12,\,-16),$$

$$\overrightarrow{PB}=B-P=(24-16,\,0-16)=(8,\,-16).$$

Using the dot-product formula for the angle between two vectors,

$$\cos\theta=\dfrac{\overrightarrow{PC}\cdot\overrightarrow{PB}} {|\overrightarrow{PC}|\;|\overrightarrow{PB}|}.$$

The dot product is

$$\overrightarrow{PC}\cdot\overrightarrow{PB}=(-12)(8)+(-16)(-16) =-96+256=160.$$

The magnitudes are

$$|\overrightarrow{PC}|=\sqrt{(-12)^{2}+(-16)^{2}} =\sqrt{144+256}=20,$$

$$|\overrightarrow{PB}|=\sqrt{8^{2}+(-16)^{2}} =\sqrt{64+256}=8\sqrt5.$$

Therefore

$$\cos\theta=\dfrac{160}{20\cdot8\sqrt5} =\dfrac{160}{160\sqrt5} =\dfrac1{\sqrt5}.$$

Next, using $$\sin^{2}\theta=1-\cos^{2}\theta$$ we get

$$\sin\theta=\sqrt{1-\dfrac1{5}} =\sqrt{\dfrac{4}{5}} =\dfrac{2}{\sqrt5}.$$

Finally,

$$\tan\theta=\dfrac{\sin\theta}{\cos\theta} =\dfrac{\dfrac{2}{\sqrt5}}{\dfrac1{\sqrt5}} =2.$$

Thus an admissible value of $$\tan\theta$$ is $$2$$.

Hence, the correct answer is Option C.