Question

What is the slope of the tangent line of `(x^2-y^2)/(sqrt(4y-x)+xy) =C`, where C is an arbitrary constant, at `(1,1)`?

Answer 1

Please see below.

Explanation:

Given that `(1,1) lies on the graph, we see that`C = 0#.

The equation is equivalent to `x^2-y^2=0` with `y >= 0`.

Differentiating implicitly, we have:

`2x-2y dy/dx = 0`, so

`dy/dx = y/x = {(1,"if",x > 0),(-1,"if",x < 0):}`

At `(1,1)`, the slope is `1`.

`(x^2-y^2)/(sqrt(4y-x)+xy) =0` is graphed below.

(Note that `x^2-y^2 = (x-y)(x+y)` so the graph is parts of `y=x` and `y=-x`.)

graph{(x^2-y^2)/(sqrt(4y-x)+xy) =0 [-6.496, 7.55, -1.824, 5.2]}