What is the slope of the tangent line of #(x-y)^3-y^2/x= C #, where C is an arbitrary constant, at #(1,-2)#?

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Answer 1 · 2016-12-27T06:23:59

#y=31/29x-89/29#

The point# P(1, -2)# lies on the curve. So,

#(1+3)^3=(-2)^2/1=23=C#

From

(x-y)^3-y^2/x=23,

y' at P =31/29.

The equation to the tangent at P is

#y+2=31/29(x-1)#, This simplifies to

#y=31/29x-89/29#

graph{((x-y)^3-y^2/x-23)(y-31/29x+89/29)=0x^2 [-5, 5, -10, 10]}