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

Question

1640 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-02T15:20:43

The slope of the tangent is $0$ $0$ (and $C$ $C$ is not "arbitrary", it is $- 3$ $-3$ ).

$(x - y) (x + y) - x y + y = C$ $(x-y)(x+y)-xy+y= C$

Instead of differentiating immediately, let's rewrite the first product.

$x^{2} - y^{2} - x y + y = C$ $x^2-y^2-xy+y=C$ .

This form is (I think) simpler to differentiate.

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

At the point $(1, 2)$ $(1,2)$ , we get

$2 (1) - 2 (2) \frac{d y}{d x} - 2 - (1) \frac{d y}{d x} + \frac{d y}{d x} = 0$ $2(1) - 2(2) dy/dx - 2 - (1) dy/dx + dy/dx = 0$

Solve for $\frac{d y}{d x}$ $dy/dx$ to see that $\frac{d y}{d x} = 0$ $dy/dx = 0$

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