Find an equation of the tangent line to the graph at the given point. ?

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1 Answer
Mar 17, 2018

#color(blue)(y=-2/11x-30/11)#

Explanation:

Expand the equation. This is not essential, but it seems easier when differentiating implicitly:

#3(x^2+y^2)^2=100(x^2+y^2)#

#3x^4+6x^2y^2+3y^4-100x^2+100y^2=0#

Differentiating implicitly:

#dy/dx(3x^4+6x^2y^2+3y^4-100x^2+100y^2)=#

#12x^3+12x^2y*dy/dx+12xy^2+12y^3*dy/dx-200x+200y*dy/dx=0#

#12x^2y*dy/dx+12y^3*dy/dx+200y*dy/dx=-12x^3-12xy^2+200x#

Factor:

#dy/dx(12x^2y+12y^3+200y)=-12x^3-12xy^2+200x#

#dy/dx=(-12x^3-12xy^2+200x)/(12x^2y+12y^3+200y)#

#dy/dx=(-3x^3-3xy^2+50x)/(3x^2y+3y^3+50y)=-((3x^3+3xy^2-50x))/(3x^2y+3y^3+50y)#

Let #m=-((3x^3+3xy^2-50x))/(3x^2y+3y^3+50y)#

Using point slope form of a line:

#y-(-2)=m(x-(-4))#

#y+2=m(x+4)#

#y=mx+4m-2#

Pugging in values for #x and y# into #bbm# i.e. #(-4,-2)#

#-((3(-4)^3+3(-4)(-2)^2-50(-4)))/(3(-4)^2(-2)+3(-2)^3+50(-2))=-2/11#

#m=-2/11#

Substituting in:

#y=mx+4m-2#

#y=-2/11x+4(-2/11)-2#

#color(blue)(y=-2/11x-30/11)#

GRAPH:

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