I'm given a curve defined by [tex]2y^3 + 6x^2y - 12x^2 + 6y = 1[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Its derivative is [tex]\frac{4x - 2xy}{x^2 + y^2 + 1}[/tex]

One of the problems is as follows: "Write an equation of each horizontal tangent line to the curve."

I reasoned that if the top of the derivative is equal to zero, the tangent line will be horizontal at that point. Therefore, I solved for [tex]4x - 2xy = 0[/tex]

[tex]4x = 2xy[/tex]

[tex]y = \frac{4x}{2x} = 2[/tex]

This is the only solution I can find, but the wording of the question, "EACH horizontal line" makes me feel as though I'm missing something and there are actually more. Is y = 2 the only one?

The next problem is as follows: "The line through the origin with slope -1 is tangent to the curve at point P. Find the x- and y-coordinates of point P."

Correct me if I'm wrong, but doesn't this mean that we have to solve for x AND y?

[tex]-1 = \frac{4x - 2xy}{x^2 + y^2 + 1}[/tex]

Am I going the wrong way about this, or is there a way to simplify this equation that I'm not seeing?

Thanks.

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# Fairly simple calculus question

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