Question

If the tangent and normal drawn to the curve $x=a(\theta+\sin \theta), y=a 91-\cos \theta)$ at $P \left(\theta=\frac{\pi}{2}\right)$ cuts the $X$-axis at $A$ and $B$ respectively, then the area (in sq. units) of $\triangle PAB$ is

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

Answer 1

Answer: (C)

Given curve is
$x=a(\theta+\sin \theta), y=a(1-\cos \theta)$
$\therefore \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}=\tan \frac{\theta}{2}$
$\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
$\Delta P A B=\frac{1}{2}\begin{vmatrix}a\left(\frac{\pi}{2}+1\right) & a & 1 \\ \frac{a \pi}{2} & 0 & 1 \\ a\left(\frac{\pi}{2}+2\right) & 0 & 1\end{vmatrix}$
$=\frac{1}{2}\left|a\left[\frac{a \pi}{2}-\frac{a \pi}{2}-2 a\right]\right|=a^{2}$