JEE Main 2022 — Mathematics Calculus

Given,

Slope of normal $=\frac{-dx}{dy}=\frac{{x}^{2}}{xy-{x}^{2}{y}^{2}-1}$=dy−dx​=xy−x2y2−1x2​

$${x}^{2}{y}^{2}dx+dx-xydx={x}^{2}dy$$x2y2dx+dx−xydx=x2dy

$${x}^{2}{y}^{2}dx+dx={x}^{2}dy+xydx$$x2y2dx+dx=x2dy+xydx

$${x}^{2}{y}^{2}dx+dx=x(xdy+ydx)$$x2y2dx+dx=x(xdy+ydx)

$${x}^{2}{y}^{2}dx+dx=xd(xy)$$x2y2dx+dx=xd(xy)

$$\frac{dx}{x}=\frac{dxy}{1+{x}^{2}{y}^{2}}$$xdx​=1+x2y2dxy​

$$\mathrm{ln}kx={\mathrm{tan}}^{-1}(xy)\cdots (i)$$lnkx=tan−1(xy)⋯(i)

Curve passes though $(1,1)$(1,1)

So, In $k=\frac{\pi }{4}\Rightarrow k={e}^{\frac{\pi }{4}}$k=4π​⇒k=e4π​

Now from equation $(i)$(i)

We get, $\frac{\pi }{4}+\mathrm{ln}x={\mathrm{tan}}^{-1}(xy)$4π​+lnx=tan−1(xy)

$$xy=\mathrm{tan}(\frac{\pi }{4}+\ell nx)$$xy=tan(4π​+ℓnx)

$$xy=(\frac{1+\mathrm{tan}(\ell nx)}{1-\mathrm{tan}(\ell nx)})\cdots (\mathrm{ii})$$xy=(1−tan(ℓnx)1+tan(ℓnx)​)⋯(ii)

Put $x=e$x=e in $(\mathrm{ii})$(ii)

We get,$e.y(e)=\frac{1+\mathrm{tan}1}{1-\mathrm{tan}1}$e.y(e)=1−tan11+tan1​