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SUMMARY

The discussion focuses on finding the equation of the tangent line to the implicit function defined by x3 + y3 - 6xy = 0 at the point (4/3, 8/3). The correct derivative is established as dy/dx = (6y - 3x2) / (3y2 - 6x), leading to a slope of m = -8/5. The final equation of the tangent line is confirmed as y - (8/3) = -(8/5)(x - (4/3)). The user initially miscalculated due to a sign error in their differentiation process.

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communitycoll
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Homework Statement


Find an equation of the tangent line to x^3 + y^3 - 6xy = 0 at the point ((4/3), (8/3))


Homework Equations


I got y = -(8/5)x + (24/5). Is this correct?


The Attempt at a Solution


Lots of algebra involved. Sorry. but I'd rather not type it. I take the derivative of each term and eventually get:

[-3x^2 - 6y] / [3y^2 - 6x].

This simplifies to:

[x^2 + 2y] / [2x - y^2].

I substitute the stuff in appropriately, I get m = -(8 / 5).

Then I do:

y - (8/3) = -(8 / 5)(x - (4/3)).

I ask because Wolfram Alpha disagrees with me:
http://www.wolframalpha.com/input/?...3+++y^3+-+6xy+=+0+at+the+point+((4/3),+(8/3))
 
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communitycoll said:
Lots of algebra involved. Sorry. but I'd rather not type it. I take the derivative of each term and eventually get:

[-3x^2 - 6y] / [3y^2 - 6x].

You're off by a sign, it should be

\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}
 
I see. I didn't distribute a "- 6" to a "+ y" properly. Thanks.
 

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