Question

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

Answer 1

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

`(x-y^2)/(xy+1)=-1` of the system

`(x-y^2)/(xy+1)=C`

passes through (1, 2).

The equation becomes

`(x-y+1)(y+1)=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]}