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

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Answer 1 · 2018-03-17T00:52:24

Please see below.

Given that $(1, 1) l i e s o n t h e g r a p h, w e s e e t ˆ$ $(1,1) lies on the graph, we see that$ C = 0#.

The equation is equivalent to $x^{2} - y^{2} = 0$ $x^2-y^2=0$ with $y \geq 0$ $y >= 0$ .

Differentiating implicitly, we have:

$2 x - 2 y \frac{d y}{d x} = 0$ $2x-2y dy/dx = 0$ , 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]}

What is the slope of the tangent line of (x^2-y^2)/(sqrt(4y-x)+xy) =C x2−y2√4y−x+xy=C (x^2-y^2)/(sqrt(4y-x)+xy) =C , where C is an arbitrary constant, at (1,1)(1,1)(1,1)?