Question

If $ST$ and $SN$ are the lengths of the subtangent and the subnormal at the point $\theta =\frac {\pi}{2}$ on the curve $x = a$ $(\theta + Sin \theta).y=a(1-Cos\theta) a \neq 1$ ,then

• A. $ST = SN$
• B. $ST=2SN$
• C. $ST^2 =aSN^3$
• D. $ST^3 =aSN$

Answer 1

Answer: (A)

Given that, $x=a(\theta+\sin \theta)$ and $y=a(1-\cos \theta)$
$\Rightarrow \frac{d x}{d \theta}=a(1+\cos \theta)$ and $\frac{d y}{d \theta}=a \sin \theta$
$\therefore \quad \frac{d y}{d x}=\frac{a \sin \theta}{a(1+\cos \theta)}$
$=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}$
$=\tan \frac{\theta}{2}$
Now, length of sub tangent $=\left|\frac{y}{d y / d x}\right|$
$\therefore \, S T=\frac{a(1-\cos \theta)}{\tan \frac{\theta}{2}}$
$=a \cdot \frac{2 \sin ^{2} \frac{\theta}{2}}{\sin \frac{\theta}{2}} \cdot \cos \frac{\theta}{2} \\
= a \sin \theta$
$\Rightarrow $ Length of subtangent at $\theta=\frac{\pi}{2}$,
$ S T=a \,\sin \frac{\pi}{2}=a $
and length of subnormal $=\left|y \frac{d y}{d x}\right| $
$\Rightarrow \, S N=a(1-\cos \theta) \cdot \tan \frac{\theta}{2}$
$\Rightarrow \,a \cdot 2 \sin ^{2} \frac{\theta}{2} \tan \frac{\theta}{2}$
$\Rightarrow $ Length of subnormal at $\theta=\frac{\pi}{2}$,
$S N=a \cdot 2 \cdot \frac{1}{2}=a$
Hence, $S N=S T$