Question

If $\mathrm{ST}$ and $\mathrm{SN}$ 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. $\mathrm{ST}=\mathrm{SN}$
• B. $\mathrm{ST}=2\mathrm{SN}$
• C. ${\mathrm{ST}}^2={\mathrm{aSN}}^3$
• D. ${\mathrm{ST}}^3=\mathrm{aSN}$

Answer 1

The correct option is A

$\mathrm{ST}=\mathrm{SN}$

Explanation for the correct option.

Step 1: Find $\frac{dy}{dx}$

Given that, $\mathrm{x}=\mathrm{a}(\theta +\sin \theta )\text{and}\mathrm{y}=\mathrm{a}(1-\cos \theta )$.

Differentiating we get:

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{d}\theta }=\mathrm{a}(1+\cos \theta )\text{and}\frac{\mathrm{dy}}{\mathrm{d}\theta }=\mathrm{asin}\theta$

$\begin{array}{rcl}& \Rightarrow & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{asin}\theta }{\mathrm{a}(1+\cos \theta )}\\ & =& \frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}\\ & =& \tan \frac{\theta }{2}\end{array}$

Step 2: Find the length of the subtangent

Now, the length of subtangent $=\left|\frac{\mathrm{y}}{\mathrm{dy}/\mathrm{dx}}\right|$
$\begin{array}{rcl}\mathrm{ST}& =& \frac{\mathrm{a}(1-\cos \theta )}{\tan \frac{\theta }{2}}\\ & =& a\frac{2{\sin }^2\frac{\theta }{2}}{\sin \frac{\theta }{2}}\cos \frac{\theta }{2}\\ & =& a\sin \theta \end{array}$
At $\theta =\frac{\pi }{2}$ ,
$\begin{array}{rcl}\mathrm{ST}& =& \mathrm{asin}\frac{\pi }{2}\\ & =& \mathrm{a}\end{array}$

Step 3: Find the length of the subnormal

Length of subnormal $=\left|\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}\right|$
$\begin{array}{rcl}\mathrm{SN}& =& \mathrm{a}(1-\cos \theta )\tan \frac{\theta }{2}\\ & =& \mathrm{a}\times 2{\sin }^2\frac{\theta }{2}\tan \frac{\theta }{2}\end{array}$
At $\theta =\frac{\pi }{2}:$

$\begin{array}{rcl}\mathrm{SN}& =& \mathrm{a}·2·\frac{1}{2}\\ & =& \mathrm{a}\dots .\left(2\right)\end{array}$

Hence, $\mathrm{SN}=\mathrm{ST}\dots .$ By (1) and (2)

Hence, option A is correct.