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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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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
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1 Answer

So, $2Y=3X^2$

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

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

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

So, $X\ne0$

$X^3=\frac{27}{16}\implies \frac{3Y^2}X=\frac{3\left(\frac{3X^2}2\right)^2}X=\frac{27}4\cdot X^3$ (as $x\ne0$)

$\implies \frac{3Y^2}X=\frac{27}4\cdot \frac{27}{16}\ne1\implies 3Y^2-X\ne0$

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

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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
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