# To find coordinate of the point, need help, please

• Sep 7th 2012, 05:16 AM
Minwoo
To 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 AM
Plato
Re: To find coordinate of the point, need help, please
Quote:

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

Use implicit differentiation to find $y^{\prime}$.
Next solve $y^{\prime}=0$.
• Sep 7th 2012, 05:53 AM
Minwoo
Re: 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 AM
Plato
Re: To find coordinate of the point, need help, please
Quote:

Originally Posted by Minwoo
Yes, I found dy/dx is (x-y)/(x-2y), then what is the next step? Thank you

At what points on the curve is it true that $\frac{dy}{dx}=0~?$
• Sep 7th 2012, 06:00 AM
Minwoo
Re: To find coordinate of the point, need help, please
Sorry, I really dunno
• Sep 7th 2012, 06:09 AM
Plato
Re: To find coordinate of the point, need help, please
Quote:

Originally Posted by Minwoo
Sorry, I really dunno

Are you saying that you cannot solve $\frac{x-y}{x-2y}=0~?$

If you are, then you have no business even trying this problem!
• Sep 7th 2012, 06:16 AM
HallsofIvy
Re: 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, $x^2 -2xy+2y^2=4$, implicit differentiation gives $2x- 2y- 2xy'z+ 4yy'= 0$. Now, because you are looking for points where y'= 0, put y'= 0 into that equation: $2x- 2y= 0$. That tells you that y= x. Replace y with x in the original equation, $x^2 -2xy+2y^2=4$ to get an equation in x only and solve that for x.
• Sep 8th 2012, 01:49 AM
Minwoo
Re: To find coordinate of the point, need help, please
Thank you for the help