Find point nearest to the origin Find the points on the curve $5x^2 - 6xy + 5y^2 = 4$ that are nearest the origin.
The first method I've tried is I've taken the derivative of the equation to optimize (Pythagorean Theorem) and also the function of the curve using implicit differentiation and plugged stuff in (the way I've done it so far). I got $y = x$ which I tried to plug into the function of the curve but didn't match the answers at the back of the textbook. I stared at it and figured I made a sign error somewhere since $-y = x$ worked but couldn't find it.
The second method I tried was solving for the function of the curve in terms of $y$ and tried both completing the square and using the quadratic formula to find $x$ which gave the same failed answer where $y$ subtracted to nonexistence when I plugged them into the function of the curve.
I don't know what lagrange multipliers are, which was the only solution on the web. I hope it could be done just with simple algebra (or calculus) since the textbook expects that (I think).
 A: Hint: Using a change of coordinates $X=x+y$, $Y= x-y$, the problem becomes 
$$ 5\frac{X^2+Y^2}{2} -6\frac{X^2-Y^2}{4} = 4,$$
or that
$$ X^2 + 4Y^2 = 4.$$
You should recognize this as an ellipse, with minor axis __ and major axis __.
Hence, the answer is __.

With regards to your differentiation solution:
$x^2 + y^2$ is minimized, when $2x + 2y \frac{dy}{dx} = 0$, or that $\frac{dy}{dx} = -\frac{y}{x}$.
In the original equation, we have $ 10x - 6y - 6x \frac{dy}{dx} + 10y \frac{dy}{dx}$, or that $\frac{dy}{dx} = \frac{6y-10x}{10y-6x}  $.
Hence, solving for $-\frac{y}{x} = \frac{dy}{dx} = \frac{6y-10x}{10y-6x}$, we get $-y(10y-6x)=x(6y-10x)$, or that $-10y^2 +6xy = 6xy - 10x^2$, so $10(x^2-y^2)=0$. This has solution set $x=y, x=-y$.
You then need to check the second order condition, to see which (if any) gives your a global minimum.
A: Let $x=r\cos\theta$ and $y=r\sin\theta$. Then
$$4=5x^2-6xy+5y^2=r^2-6r^2\cos\theta\sin\theta=r^2(5-3\sin 2\theta).$$
To minimize $r^2$, we maximize $5-3\sin 2\theta$. The maximum value of $5-3\sin 2\theta$ is $8$. 
A: You have to solve the following optimization problem:
Min $x^2+y^2$
Subject to:
$5x^2 -6xy+5y^2=4$
You could solve it using calculus as follows. Use the constraint to express $y$ in terms of $x$ and substitute that in the objective function. Then use the usual method of finding the optimal value by setting the $f'(x)=0$ and checking that $f''(x) >0$.
