Question

If ST and SN are the lengths of the subtangent and the subnormal at the point $\theta =\frac{\pi }{2}$ on the curve $x=a(\theta +sin\theta ),y=at(1-cos\theta ),a\ne 1,$ then
[Karnataka CET 2005]

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

Answer 1

The correct option is A ST = SN

$y=(x+\sqrt{1+x^2}{)}^n\Rightarrow \frac{dy}{dx}=n(x+\sqrt{1+x^2}{)}^{n-1}(1+\frac{x}{\sqrt{1+x^2}})$
$\Rightarrow \frac{dy}{dx}=\frac{n(x+\sqrt{1+x^2}{)}^n}{\sqrt{1+x^2}}$
$\Rightarrow (\sqrt{1+x^2}\frac{dy}{dx}=n(x+\sqrt{1+x^2}{)}^n$
$\Rightarrow \frac{d^{2}y}{d{x}^2}.\sqrt{1+x^2}+\frac{dy}{dx}\left(\frac{x}{\sqrt{1+x^2}}\right)$
$=n^2{(x+\sqrt{1+x^2})}^{n-1}(1+\frac{x}{\sqrt{1+x^2}})$
$\Rightarrow (1+x^2).\frac{d^{2}y}{d{x}^2}+x.\frac{dy}{dx}=n^2(x+\sqrt{1+x^2}{)}^n$
$\Rightarrow (1+x^2)\frac{d^{2}y}{d{x}^2}+x.\frac{dy}{dx}=n^{2}y.$