# find the points on a curve where the tangent line is horizonta

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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Just to add a clarification to lab bhattacharjee's answer, a fraction is equal to 0 when the numerator is equal to 0. So you get $2Y-3X^2=0$ and then ultimately $2Y=3X^2$ (the first line of his answer). –  Conan Wong Dec 13 '12 at 16:29
that's clear....but what would be the point (s)?, the prof. hinted that the point(s) on the curve most satisfy the equation for the curve x3+y3=2xy –  carolina nunez Dec 13 '12 at 16:59
Keep in mind that this curve is not the graph of a function: some values of $x$ produce more than one value of $y$. The point $(0,0)$ is on the curve, but makes both the numerator and the denominator in the derivative of the implicit function equal zero. [Hint: on such a curve, sometimes a point can have two slopes!] –  RecklessReckoner Apr 28 '13 at 6:25
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. –  RecklessReckoner May 11 '14 at 3:36

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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the points have to satisfy the equation of the curve.....i understand what you did, just don't know how to related it to the question –  carolina nunez Dec 13 '12 at 16:44
You have $2y=3x^2$. The curve is $x^3+y^3=2xy$. Multiply by $8$ to get $8x^3+(2y)^3=8x(2y)$. Use $2y=3x^2$ to rewrite this as $8x^3+27x^6=24x^3$. Can you take it from there? –  Gerry Myerson Dec 14 '12 at 0:24
But beware: x = 0 also solves that last equation, but does not seem to work in the derivative equation. There is something "funny" going on at the origin -- you'll need to check the curve to see why. –  RecklessReckoner Apr 28 '13 at 1:13
@RecklessReckoner, thanks for input. The values of $X$ deserves validation. –  lab bhattacharjee Apr 28 '13 at 6:08

[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:

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.

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