Finding the largest coordinate of $y$ given the ellipse $x^2+y^2-xy=1.$ The equation of the ellipse is $x^2+y^2-xy=1$.
I am asked to find the the coordinate of the point $p$ on the ellipse with largest $y$ value.
Thanks in advance!
So I have that $\frac{dy}{dx}=\frac{(-2x+y)}{(2y-x)}$, but I don't know how to proceed from here.
 A: We complete the square: 
$$x^2+y^2-xy=\frac{3}{4}y^2+\left(x-\frac {y}{2}\right)^2.$$
So to maximize $y$ given $x^2+y^2-xy=1$ we should set $x=\frac{y}{2}$. That gives $y=\frac{2}{\sqrt{3}}$.
A: The "critical" $y$-values will happen where $x$ can/cannot be solved for; that is, where the discriminant of the quadratic equation in $x$ is equal to $0$ exactly. This means that $$\Delta=b^2-4ac=y^2-4(1)(y^2-1)=-3y^2+4=0$$ which gives the maximum $y$-value as $$y^2=\frac{-4}{-3}\implies y=\frac{2}{\sqrt3}$$
A: You can always use Lagrange multipliers, maximizing $y$ subject to the constraint $x^2 + y^2 - xy = 1$. The result is
$$0 = \lambda(2x - y)$$
$$1 = \lambda(2y - x)$$
 $\lambda \neq 0$ due to the second equation, so we can divide the first equation by $\lambda$ to get $y = 2x$. Plugging this into the constraint gives $3x^2 = 1$ or $x = \pm{1 \over \sqrt{3}}$. Thus $y = 2x = \pm{2 \over \sqrt{3}}$. The plus sign gives the maximum and the minus gives the minimum.
A: You could differentiate implicitly, solve for $\frac{dy}{dx}$, and treat it as a maximization problem.
Might be an easier way to do this though, that's just my first thought.
A: Consider the equation $$y^2-xy+x^2-1=0$$ as a quadratic equation in $y$. Differentiating with respect to $y$ gives $$2y-x=0$$ or $$y=\frac{x}{2}$$ Now put the value in the equation and find $x$. (So you have $y$ as well.)
A: defferentiate the relation. When $y$ is max $\dfrac{dy}{dx}=0$. 
