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

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Answer 1 · 2016-02-02T15:20:43

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

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

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

#x^2-y^2-xy+y=C#.

This form is (I think) simpler to differentiate.

#2x-2y dy/dx - y - x dy/dx + dy/dx = 0#

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

#2(1) - 2(2) dy/dx - 2 - (1) dy/dx + dy/dx = 0#

Solve for #dy/dx# to see that #dy/dx = 0#