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

Jan 15, 2017

Only a pair of straight lines form the graph. The slope of the the line through (1, 3) is 1.

Explanation:

For the parameter value C = -1, the member

$\frac{x - {y}^{2}}{x y + 1} = - 1$ of the system

$\frac{x - {y}^{2}}{x y + 1} = C$

passes through (1, 2).

The equation becomes

$\left(x - y + 1\right) \left(y + 1\right) = 0$, giving a pair of straight lines.

The slope of the first line #x-y+1 = 9 that passes throgh (1, 2) is 1.

See graphical illustration

Note:

By choosing a point (2, 1), instead, C becomes 1/3, giving a

hyperbola.

For, the choice (2, 1), it is the parabola ${y}^{2} = x$.

graph{(x-y^2)/(xy+1)+1=0 [-10, 10, -5, 5]}