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

Question

The curve is given by $x^{2} + 3 y^{2} = 1 + 3 x y$ $x^2+3y^2=1+3xy$

and $\frac{d y}{d x} = \frac{3 y - 2 x}{6 y - 3 x}$ $(dy)/(dx)=(3y-2x)/(6y-3x)$

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Answer 1 · 2018-04-05T06:43:27

$(1, 2) and (- 1, - 2)$ $(1,2) and (-1,-2)$ 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 $m =$ $m=$ the slope of the tangent line at that point.

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

$m = \pm \infty$ $m=+-oo$ means the tangent line is vertical at that point.

$\frac{d y}{d x} = \frac{3 y - 2 x}{6 y - 3 x} = \pm \infty$ $dy/dx=(3y-2x)/(6y-3x)=+-oo$

$6 y - 3 x = 0$ $6y-3x=0$

$6 y = 3 x$ $6y=3x$

$x = 2 y$ $x=2y$

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

$x^{2} + 3 y^{2} = 1 + 3 x y$ $x^2+3y^2=1+3xy$

$4 y^{2} + 3 y^{2} = 1 + 6 y^{2}$ $4y^2+3y^2=1+6y^2$

$y^{2} = 1$ $y^2=1$

$y = \pm 1$ $y=+-1$

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

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

$(2,1) and (-2,-1)$ 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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