JEE Main 2015 — Application of Derivatives Question with Solution

At $t=\frac{\pi }{4}$,

$x=2\frac{1}{\sqrt{2}}+2\frac{\pi }{4}.\frac{1}{\sqrt{2}}=\left(\sqrt{2}+\frac{\pi }{2\sqrt{2}}\right)=\left(\frac{4+\pi }{2\sqrt{2}}\right)$

$y=2\frac{1}{\sqrt{2}}-2\frac{\pi }{4} \cdot \frac{1}{\sqrt{2}} = \left(\sqrt{2}- \frac{\pi }{2\sqrt{2}}\right)=\left(\frac{4-\pi }{2\sqrt{2}}\right)$

$\frac{dy}{dx}=2\cos t-2 \left[\cos t+t (-\sin t)\right]=2t\sin t$

$\frac{dx}{dt}= -2\sin t+2 \left[\sin t+t\cdot \cos t\right]=2t\cos t$

$\frac{dy}{dx}=\tan t$  at $\mathrm{t}=\frac{\mathrm{\pi }}{4}\mathrm{ }\Rightarrow \mathrm{ }\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=\mathrm{tan }\frac{\mathrm{\pi }}{4}=1$

$\frac{dy}{dx}=1$ .

Slope of tangent is $1$ & therefore slope of normal would be $-1$.  

Equation of normal  $y- \left(\frac{4-\pi }{2\sqrt{2}}\right)= -1 \left(x- \left(\frac{4+\pi }{2\sqrt{2}}\right)\right)$

$x+y=\left(\frac{4+\pi }{2\sqrt{2}}\right)+ \left(\frac{4-\pi }{2\sqrt{2}}\right)$

$\Rightarrow x+y=\frac{8}{2\sqrt{2}}$

Distance from origin $= \left|\frac{\frac{-8}{2\sqrt{2}}}{\sqrt{2}}\right|=2$