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

Jan 12, 2017

$3 x - 5 y + 14 = 0$. Tangent-inclusive graph is inserted.

#### Explanation:

As P(-3, 1) is on the graph,

C = ((-3)(1)-1^2)(1-3)=0.

Differentiating,

$\left(x y ' + y - 2 y y '\right) \left(1 + x\right) + \left(x y - {y}^{2}\right) \left(1\right) = 0$.

At P, $\left(- 3 y ' + 1 - 2 y '\right) \left(1 - 3\right) + \left(- 3 - 1\right) = 0$, giving the slope of the tangent at P

$y ' = \frac{3}{5}$. So, the equation of the tangent is

$y - 1 = \frac{3}{5} \left(x + 3\right)$, giving

$3 x - 5 y + 14 = 0$.

I have used a parallel line, in proximity of the tangent, to keep off the

gap, at the point of contact. In this graphics method, P appears as a

gap, for the exact equation of the tangent. For the interested reader,

this graph is also included. In this graph, the pixels at P do not glow.

graph{((xy-y^2)(1+x)-8)(3x-5y+13.7)=0 [-9.79, 9.785, -4.895, 4.895]}

graph{((xy-y^2)(1+x)-8)(3x-5y+14)=0 [-9.79, 9.785, -4.895, 4.895]}