# What is the slope of the tangent line of (y/x)e^(x/y-x^2)= C , where C is an arbitrary constant, at (1,2)?

Jun 2, 2018

10

#### Explanation:

Taking the logarithm of both sides gives us

$\ln y - \ln x + \frac{x}{y} - {x}^{2} = \ln C$

Differentiating with respect to $x$ yields

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{1}{x} + \frac{1}{y} - \frac{x}{y} ^ 2 \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x = 0 \implies$

$\left(\frac{1}{y} - \frac{x}{y} ^ 2\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x + \frac{1}{x} - \frac{1}{y} \implies$

$x \left(y - x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 {x}^{2} + 1\right) {y}^{2} - x y$

At the point $\left(1 , 2\right)$ we have

$1 \times \left(2 - 1\right) \times \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 \times {1}^{2} + 1\right) \times {2}^{2} - 1 \times 2 \implies$
$\frac{\mathrm{dy}}{\mathrm{dx}} = 10$

Hence the slope of the tangent at $\left(1 , 2\right)$ is 10

Note

I have assumed that tangent at $\left(1 , 2\right)$ means tangent to the curve at $\left(1 , 2\right)$. Of course, this means that the constant $C$ can not be arbitrary - but rather has to be $\frac{2}{\sqrt{e}}$