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I want to find the coordinates where the tangent to the curve is parallel to the x and y axis.

The curve is $$2x^2 +xy - y^2 +18 = 0 $$

$$ Dy/dx = (-4x-y)/(x-2y) $$

Am I correct in saying that to work out the tangent parallel to x axis, I would say that -4x-y=0 as the tangent would be horizontal and so the change in y has to be zero as therefore $$0=0/(x-2y). $$

Subsequently, to work out the tangent to the y axis, is it x-2y=0 as the change in x would be 0 as the tangent would be vertical and so $$0=-4x-y$$

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For horizontal tangent, set the numerator of $dy/dx$ equal to zero.

For vertical tangent, set the denominator of $dy/dx$ equal to zero.

In the case where both equal zero, you have neither a horizontal tangent nor a vertical tangent.

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Considering the implicit function $$f(x,y)=2x^2 +xy - y^2 +18 = 0$$ you properly found that $$\frac{dy}{dx}=-\frac{4 x+y}{x-2 y}$$ So, the tangent is horizontal if $y=-4x$ and for this condition $$f(x,-4x)=18-18x^2=0$$ then $x=\pm 1$ and the corresponding $y$'s.

Similarly, the tangent is vertical if $y=\frac x 2$ and for this condition $$f(x,\frac x 2)=\frac{9 x^2}{4}+18=0$$ which does not show real solutions.

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