Find the coordinates of the points on the curve X^2 -2xy+2y^2=4 at which the tangent is parallel to x-axis.

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- Sep 7th 2012, 05:16 AMMinwooTo find coordinate of the point, need help, please
Find the coordinates of the points on the curve X^2 -2xy+2y^2=4 at which the tangent is parallel to x-axis.

- Sep 7th 2012, 05:36 AMPlatoRe: To find coordinate of the point, need help, please
- Sep 7th 2012, 05:53 AMMinwooRe: To find coordinate of the point, need help, please
Yes, I found dy/dx is (x-y)/(x-2y), then what is the next step? Thank you

- Sep 7th 2012, 05:56 AMPlatoRe: To find coordinate of the point, need help, please
- Sep 7th 2012, 06:00 AMMinwooRe: To find coordinate of the point, need help, please
Sorry, I really dunno

- Sep 7th 2012, 06:09 AMPlatoRe: To find coordinate of the point, need help, please
- Sep 7th 2012, 06:16 AMHallsofIvyRe: To find coordinate of the point, need help, please
You don't really

**have**to find the derivative explicitely. Going back to your original equation, $\displaystyle x^2 -2xy+2y^2=4$, implicit differentiation gives $\displaystyle 2x- 2y- 2xy'z+ 4yy'= 0$. Now, because you are looking for points where y'= 0,**put y'= 0**into**that**equation: $\displaystyle 2x- 2y= 0$. That tells you that y= x. Replace y with x in the original equation, $\displaystyle x^2 -2xy+2y^2=4$ to get an equation in x only and solve that for x. - Sep 8th 2012, 01:49 AMMinwooRe: To find coordinate of the point, need help, please
Thank you for the help