Question

The slope of normal at any point $(x,y),x>0,y>0$ on the curve $y=y(x)$ is given by $\frac{x^2}{xy-x^2{y}^2-1}$. If the curve passes through the point $(1,1)$, then $e.y(e)$ is equal to

• A. $\frac{1-\tan (1)}{1+\tan (1)}$
• B. $\tan (1)$
• C. 1
• D. $\frac{1+\tan (1)}{1-\tan (1)}$

Answer 1

Answer: (D)

Slope of normal $=\frac{-dx}{dy}=\frac{x^2}{xy-x^2{y}^2-1}$
$x^2{y}^{2}dx+dx-xydx=x^{2}dy$
$x^2{y}^{2}dx+dx=x^{2}dy+xydx$
$x^2{y}^{2}dx+dx=x(xdy+ydx)$
$x^2{y}^{2}dx+dx=xd(xy)$
$\frac{dx}{x}=\frac{d(xy)}{1+x^2{y}^2}$
$\ln kx={\tan }^{-1}(xy)\dots$ (i)
passes though $(1,1)$
$\ln k=\frac{\pi }{4}\Rightarrow k=e^{\frac{\pi }{4}}$
equation (i) be becomes
$\frac{\pi }{4}+\ln x={\tan }^{-1}(xy)$
$xy=\tan \left(\frac{\pi }{4}+\ell \ln x\right)$
$xy=\left(\frac{1+\tan (\ell nx)}{1-\tan (\ell nx)}\right)\dots$(ii)
put $x=e$ in (ii)
$\therefore \text{ ey }(e)=\frac{1+\tan 1}{1-\tan 1}$