Are There Multiple Horizontal Tangent Lines to the Curve?

[Coldie]

I'm given a curve defined by $$2y^3 + 6x^2y - 12x^2 + 6y = 1$$

Its derivative is $$\frac{4x - 2xy}{x^2 + y^2 + 1}$$

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 $$4x - 2xy = 0$$

$$4x = 2xy$$

$$y = \frac{4x}{2x} = 2$$

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?

$$-1 = \frac{4x - 2xy}{x^2 + y^2 + 1}$$

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

Thanks.

 

[mathwonk]

what about the subtle solution, x = 0? 2y^3 + 6y = 1

 

[Coldie]

Is that saying that the CURVE has a slope of -1 as it passes through the origin? I interpreted it as saying that a line drawn through the origin with a slope of -1 would be parallel to the curve itself at point P...

 

[shmoe]

I interpreted it as saying that a line drawn through the origin with a slope of -1 would be parallel to the curve itself at point P...

It's not just parallel, it's tangent...

 

[Coldie]

Oh, I see. Wouldn't x and y be zero then, though?

 

[shmoe]

Oh, I see. Wouldn't x and y be zero then, though?

Why do you say this?

That it's a tangent simplifies things greatly. You knew that at P the slope had to be -1, but you also know that P lies on the line through the origin with slope -1 as well, so you can use this to simplify $$-1 = (4x - 2xy)(x^2 + y^2 + 1)^{-1}$$ by removing either x or y.

 

[HallsofIvy]

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.

The line through the origin with slope -1 is simply y= -x. You need to find a point P= (x,y) on the curve so that the derivative is -1 and y= -x.