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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Fairly simple calculus question

**Physics Forums | Science Articles, Homework Help, Discussion**