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

##### 1 Answer
Feb 1, 2017

The slope of the tangent is $0$ at $\left(- 2 , 1\right)$, But $\left(- 2 , 1\right)$ only lies on the curve in the specific case of $C = 0$

#### Explanation:

We have:

$\left(x y - \frac{y}{x}\right) \left(x y - \frac{x}{y}\right) = C$

At $\left(- 2 , 1\right)$ we have:

$\left(- 2 + \frac{1}{2}\right) \left(- 2 + 2\right) = C \implies C = 0$

Multiplying out gives:

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

Differentiating wrt $x$ (Implicitly and using the product rule):

$\left({x}^{2}\right) \left(2 y \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(2 x\right) \left({y}^{2}\right) - 2 x - 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} - x - y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
$\therefore \left({x}^{2} y - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + x {y}^{2} - x = 0$
$\therefore \left({x}^{2} y - y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = x - x {y}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - x {y}^{2}}{{x}^{2} y - y}$

So at $\left(- 2 , 1\right)$ we have:

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 - \left(- 2\right)}{4 - 1} = 0$

So the slope of the tangent is $0$ at $\left(- 2 , 1\right)$, But $\left(- 2 , 1\right)$ only lies on the curve in the specific case of $C = 0$

The curve of this family are quote interesting:

C=3 C=1 C=0.5 C=0.25 C=0 - The specific curve that (-2,1) lies on C=-1 C=-2 