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Find the points on the curve $x^3 + y^3 = 2xy$ where the line tangent to the curve will be horizontal

I know that that this means that the derivative of the curve will be equal to 0. This is what I get: $\frac{(2Y-3X^2)}{(3Y^2-X)} = 0$ ..and then I'm stuck. Please help.

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    On reviewing this problem, there is a (small) error in the expression for the slope of the tangent line that will affect the numerical result. This does not alter certain general conclusions, however.2014-05-11

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So, $2Y=3X^2$

Putting the value of $Y$ in the given equation, $X^3(27X^3-16)=0\implies X=0,$ or $X^3=\frac{16}{27}$

But, $X=0\implies Y=0,$ from $2Y=3X^2$

$ \frac{2Y-3X^2}{3Y^2-2X}$ will be $\frac00$ hence undefined.

So, $X\ne0$

$X^3=\frac{16}{27}\implies \frac{3Y^2}{2X}=\frac{3\left(\frac{3X^2}2\right)^2}{2X}=\frac{27}{8}\cdot X^3$ (as $x\ne0$)

$\implies \frac{3Y^2}{2X}=\frac{27}{8}\cdot \frac{16}{27}\ne1\implies 3Y^2-X\ne0$

So, $ \frac{2Y-3X^2}{3Y^2-2X}$ will be $0$ as $2Y=3X^2$ but $3Y^2-2X\ne0$

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    @RecklessReckoner, thanks for input. The values of $X$ deserves validation.2013-04-28
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[just to return to this concerning a curious feature of this curve]

The curve in question is the "folium of Descartes", so called as Descartes sent this curve to Fermat as a challenge to the latter's claim for a method of determining slopes of tangent lines. Fermat handled it successfully, something made a bit more impressive by the fact that he accomplished it in 1638, prior to the more formal development of (infinitesimal) calculus by Newton and Leibniz. It looks like this:

enter image description here

I have marked horizontal tangent lines in green and vertical ones in red. I will not reiterate the work already discussed by lab bhattacharjee , but I did want to say something further about the point I raised in the comments.

We find the expression for the first derivative of the implicit functions described by the curve to be

$ y \ ' \ = \ \frac{3x^2 \ - \ 2y}{2x \ - \ 3y^2} \ \ . $

(There is an error in earlier appearances of this rational function, some now corrected.)

As we've already seen this produces horizontal tangents where $ \ y \ = \ \frac{3}{2} x^2 \ $ . The complication arises when we look for vertical tangents, which occur where $ \ y \ ' \ $ is undefined, that being $ \ x \ = \ \frac{3}{2} y^2 \ $ . (This similarity of the two equations is due to the symmetry of the folium about the line $ \ y \ = \ x \ $ . ) Putting these two equations together gives us two solutions: $ \ ( \ 0,0 \ ) \ $ and $ \ ( \ \frac{2}{3} , \frac{2}{3} \ ) . $ Only the first of these, however, corresponds to a point on the curve (the latter is not a solution to the equation for the folium).

So the origin is a point which has both a horizontal and a vertical tangent. This is a situation which can arise for self-intersecting (non-simple) curves. Finding an "indeterminate" value for $ \ y \ ' \ $ at a point is a sign that the curve has such a self-intersection.