Question

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

• A. $\frac{4}{3}$
• B. $\frac{1}{2}$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$


Equation of tangent $\left(PA\right)$ at $P(16,16)$ is
$y\left(16\right)=16\times \frac{x+16}{2}$
$\Rightarrow 2y-x=16$
$\therefore A(-16,0)$
and equation of normal $\left(PB\right)$ at $P(16,16)$ is
$y-16=-2\times (x-16)$
$\Rightarrow y-16=-2x+32$
$\Rightarrow y+2x=48$
$\therefore B(24,0)$
$PAB$ is a right-angled triangle with the right angle at $P.$
So, the circle circumscribing right-angled triangle $APB$ will have hypotenuse $AB$ as diameter.
Since $C$ is the centre of the circle, so $C$ will be the midpoint of $AB.$
$C=(\frac{-16+24}{2},0)=(4,0)$

$\therefore m_{CP}=\frac{16}{12}=\frac{4}{3},m_{PB}=-2$
Now, $\tan \theta =\left|\frac{\frac{4}{3}+2}{1-\frac{8}{3}}\right|=2$