How to find the coordinates of each point on the curve where the tangent line is vertical?

Question

The curve is given by #x^2+3y^2=1+3xy#

and #(dy)/(dx)=(3y-2x)/(6y-3x)#

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Answer 1 · 2018-04-05T06:43:27

#(1,2) and (-1,-2)# are the points where the function has vertical tangents

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We evaluate the derivative of the function at the point of tangency to find #m=#the slope of the tangent line at that point.

#m=0# means the tangent line is horizontal at that point

#m=+-oo# means the tangent line is vertical at that point.

#dy/dx=(3y-2x)/(6y-3x)=+-oo#

#6y-3x=0#

#6y=3x#

#x=2y#

We plug this into the function to solve for one coordinate of the point:

#x^2+3y^2=1+3xy#

#4y^2+3y^2=1+6y^2#

#y^2=1#

#y=+-1#

#y=1, :. x=2y=2#

#y=-1, :. x=2y=-2#

#(2,1) and (-2,-1)# are the points where the function has vertical tangents

The graphs of the function and its vertical tangents are shown below:

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