Find the coordinates of the points where the line tangent to the curve $$x^2-2xy+2y^2=4$$ is parallel to the $x$-axis, given that $$\frac{dy}{dx}=\frac{y-x}{2y-x}$$
By letting $dy/dx = 0$ I get $y=x$ which is no help… what do I do?
Thanks
[Math] Find coordinates where tangent is parallel to $x$-axis given formula for gradient/derivative
calculusimplicit-differentiation
Best Answer
All points on the curve obey the equation $x^2 - 2xy + 2y^2 = 4$. Having $y=x$ when $\frac{dy}{dx} = 0$ reduces this to $x^2 - 2x^2 + 2x^2 = 4$ or $x^2= 4$ or $x=2$ or $x=-2$, which you can find the possible $y$-coordinates for by using the equation again (substitute $x=2$ and solve for $y$, then the same for $x=-2$) so you do get your points that way. Don't forget the base equation after you compute differentials!