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

1 Answer
Cesareo R.
Jun 9, 2016

The tangent line is #y = -1 - 18/7 (x-1)#

Explanation:

Given #f(x,y,C)=0# if #p_0 = {1,-1} in f(x,y,C)=0-> f(1,-1,C)=0->C=1/2#
Now the tangent space to #f(x,y,1/2)=0# is given by

#(dy)/(dx)=-(f_x)/(f_y) =y (1/x + (2 x^7 - 2 x^2 y - 4 x^5 y + y^2 + 2 x^3 y^2)/( x^8 - 2 x^6 y + x (2 x^3-1) y^2 - 2 x^2 y^3 + y^4))#

evaluated at #p_0# gives #((dy)/(dx))_0=-18/7# and the tangent line is given by

#y = -1 - 18/7 (x-1)#

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