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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