# What is the slope of the tangent line of 1/(x^2-y)-y/x+x^3/y = C , where C is an arbitrary constant, at (1,-1)?

Jun 9, 2016

The tangent line is $y = - 1 - \frac{18}{7} \left(x - 1\right)$

#### Explanation:

Given $f \left(x , y , C\right) = 0$ if ${p}_{0} = \left\{1 , - 1\right\} \in f \left(x , y , C\right) = 0 \to f \left(1 , - 1 , C\right) = 0 \to C = \frac{1}{2}$
Now the tangent space to $f \left(x , y , \frac{1}{2}\right) = 0$ is given by

(dy)/(dx)=-(f_x)/(f_y) =y (1/x + (2 x^7 - 2 x^2 y - 4 x^5 y + y^2 + 2 x^3 y^2)/( x^8 - 2 x^6 y + x (2 x^3-1) y^2 - 2 x^2 y^3 + y^4))

evaluated at ${p}_{0}$ gives ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{0} = - \frac{18}{7}$ and the tangent line is given by

$y = - 1 - \frac{18}{7} \left(x - 1\right)$