Question

If the tangent and normal drawn to the curve $x=a(\theta +\sin \theta ),y=a91-\cos \theta )$ at $P\left(\theta =\frac{\pi }{2}\right)$ cuts the $X$-axis at $A$ and $B$ respectively, then the area (in sq. units) of $△PAB$ is

• A. $\frac{a^2}{\sqrt{2}}$
• B. $\frac{\sqrt{2}}{a^2}$
• C. $a^2$
• D. $2a^2$

Answer 1

Answer: (C)

Given curve is
$x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$
$\therefore \frac{dy}{dx}=\frac{\sin \theta }{1+\cos \theta }=\tan \frac{\theta }{2}$
$\therefore$ Slope of tangent and normal drawn to the given curve at $P\left(\theta =\frac{\pi }{2}\right)$ is
$m_T=1$ and $m_N=-1$ respectively.
Now, equation of tangent to the given curve at point $P\left(a\left(\frac{\pi }{2}+1\right),a\right)$ is
$y-a=1\left[x-a\left(\frac{\pi }{2}+1\right)\right]$
$\therefore$ Point $A$ is $\left(\frac{a\pi }{2},0\right)$
and equation of normal to the given curve at point
$P$ is $y-a=-1\left[x-a\left(\frac{\pi }{2}+1\right)\right]$
So, point $B$ is $\left(a\left(\frac{\pi }{2}+2\right),0\right)$
Therefore area of
$\Delta PAB=\frac{1}{2}\left|\begin{array}{ccc}a\left(\frac{\pi }{2}+1\right)& a& 1\\ \frac{a\pi }{2}& 0& 1\\ a\left(\frac{\pi }{2}+2\right)& 0& 1\end{array}\right|$
$=\frac{1}{2}\left|a\left[\frac{a\pi }{2}-\frac{a\pi }{2}-2a\right]\right|=a^2$