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

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Answer 1 · 2017-01-12T12:21:42

#3x-5y+14=0#. Tangent-inclusive graph is inserted.

As# P(-3, 1) is on the graph,

C = ((-3)(1)-1^2)(1-3)=0#.

Differentiating,

#(xy'+y-2yy')(1+x)+(xy-y^2)(1)=0#.

At P, #(-3y'+1-2y')(1-3)+(-3-1)=0#, giving the slope of the tangent at P

#y'=3/5#. So, the equation of the tangent is

#y-1=3/5(x+3)#, giving

#3x-5y+14=0#.

I have used a parallel line, in proximity of the tangent, to keep off the

gap, at the point of contact. In this graphics method, P appears as a

gap, for the exact equation of the tangent. For the interested reader,

this graph is also included. In this graph, the pixels at P do not glow.

graph{((xy-y^2)(1+x)-8)(3x-5y+13.7)=0 [-9.79, 9.785, -4.895, 4.895]}

graph{((xy-y^2)(1+x)-8)(3x-5y+14)=0 [-9.79, 9.785, -4.895, 4.895]}