The slope of normal at (x,y), x,y0, on y=y(x) is x^2xy-x^2y^2-1.

Question

The slope of normal at $(x,y)$, $x,y>0$, on $y=y(x)$ is $\dfrac{x^2}{xy-x^2y^2-1}$. If the curve passes through $(1,1)$, then $e\cdot y(e)$ is:

Accepted Answer

Working through the Bernoulli-type substitution and using $(1,1)$, one obtains
$$e\cdot y(e) = \frac{1+\tan(1)}{1-\tan(1)}.$$

Solutions · 1