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

Dec 27, 2016

$y = \frac{31}{29} x - \frac{89}{29}$

#### Explanation:

The point$P \left(1 , - 2\right)$ lies on the curve. So,

${\left(1 + 3\right)}^{3} = {\left(- 2\right)}^{2} / 1 = 23 = C$

From

(x-y)^3-y^2/x=23,

y' at P =31/29.

The equation to the tangent at P is

$y + 2 = \frac{31}{29} \left(x - 1\right)$, This simplifies to

$y = \frac{31}{29} x - \frac{89}{29}$

graph{((x-y)^3-y^2/x-23)(y-31/29x+89/29)=0x^2 [-5, 5, -10, 10]}