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

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Answer 1 · 2016-12-27T06:23:59

$y = \frac{31}{29} x - \frac{89}{29}$ $y=31/29x-89/29$

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

${(1 + 3)}^{3} = \frac{{(- 2)}^{2}}{1} = 23 = C$ $(1+3)^3=(-2)^2/1=23=C$

From

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

y' at P =31/29.

The equation to the tangent at P is

$y + 2 = \frac{31}{29} (x - 1)$ $y+2=31/29(x-1)$ , This simplifies to

$y = \frac{31}{29} x - \frac{89}{29}$ $y=31/29x-89/29$

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

What is the slope of the tangent line of (x-y)^3-y^2/x= C (x−y)3−y2x=C(x-y)^3-y^2/x= C , where C is an arbitrary constant, at (1,-2)(1,−2)(1,-2)?