Question

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

• A. $ST=SN$
• B. $ST=2SN$
• C. $ST2=aS{N}^3$
• D. $S{T}^3=aSN$

Answer 1

The correct option is A

$ST=SN$

Explanation for the correct option:

Step 1:Differentiate the equations presented in relation to $\theta$

$\begin{array}{rcl}\because x& =& a(\theta +\sin \theta )\\ \frac{dx}{d\theta }& =& a(1+\cos \theta )\\ \therefore y& =& a(1-\cos \theta )\\ \frac{dy}{d\theta }& =& a\sin \theta \end{array}$

Step 2: To get the derivative of $y$ relative to $x$, divide the derivatives.

$$\begin{gathered} \begin{array}{c}\therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{d\theta }}}{\frac{dx}{d\theta }}\\ =\frac{a\sin \theta }{a[1+\cos \theta ]}\\ =\left[\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}\right]\left[\because \sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\text{and}\cos \theta =2{\cos }^2\frac{\theta }{2}-1\right]\end{array} \\ =\tan \frac{\theta }{2} \end{gathered}$$

Length of subtangent at $\theta =\frac{\pi }{2}$

$\begin{array}{c}\therefore ST=a\sin \left(\frac{\pi }{2}\right)\\ =a\end{array}\left[\because \sin (90°)=\sin \left(\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right)=1\right]$

Step 3: Write the subnormal's length at $\theta =\frac{\pi }{2}$

$\text{Length of subnormal}=\left|y\left(\frac{dy}{dx}\right)\right|$

$\begin{array}{c}\text{∴SN}=a(1-\cos \theta )\tan \left(\frac{\theta }{2}\right)\\ =a\times 2{\sin }^2\left(\frac{\theta }{2}\right)\times \tan \left(\frac{\theta }{2}\right)\end{array}$

$\begin{array}{l}\text{Length of subnormal at}\theta =\frac{\pi }{2}\end{array}$

$\begin{array}{c}\text{∴SN}=2\times \left(\frac{1}{2}\right)\times a\\ =a\end{array}$

Therefore, subtangent and subnormal values are compared.

$ST=SN$

Hence, option(a) is correct.