Question #627ac

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Answer 1 · 2017-11-15T20:55:34

#m=-1/2#

we take

#x^3y^3=8#

taking out cube root

#xy=2#

isolate the variable y

#y=2/x#

deriving to find the slope of the line tangent to the curve

#y'=-2/x^2#

on the point #(2,1) -> x=2#

substitute in the derivative

#y'=-2/(2^2)#

#y'=-1/2#

the tangent line will be given by the point-slope equation

#y-y_o=m(x-x_o)#

#y-1=-1/2(x-2)#

operating

#x+2y-4=0#