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

Answer 1

The correct option is D $2$

$PAB$ 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 as we know

Since C is the center of the circle, so C will be the midpoint of $AB$.

Now given curve is $y^2=16x$

$\Rightarrow 2y\frac{dy}{dx}=16$

${\left|\frac{dy}{dx}\right|}_{(16,16)}=\frac{8}{16}=\frac{1}{2}$

Equation of $PA:(y-16)=\frac{1}{2}(x-16)$

$\Rightarrow 2y-32=x-16$

$\Rightarrow -\frac{x}{16}+\frac{y}{8}=1$

Equation of $PB:(y-16)=-2(x-16)$

$\Rightarrow 2x+y=48$

$\Rightarrow \frac{x}{24}+\frac{y}{48}=1$

So, $A=(-16,0),B=(24,0)$

$C=(\frac{-16+24}{2},0)=(4,0)$

So, slope of $CP$ is $\frac{16-0}{16-4}=\frac{4}{3}$

We know, the slope of $PB=-2$

$m_1=\frac{4}{3},m_2=-2$

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$=\left|\frac{\frac{4}{3}+2}{1-\frac{8}{3}}\right|=\left|\frac{4+6}{3-8}\right|=2$

Hence, $\tan \theta =2$