Implicit Differentiation: Multiple Solutions? Question: 
Find the coordinates of those points on the curve given by the equation:
$\ x^2-0.25xy+y^2=16 $
at which the tangent line has slope 1.
Implicitly derived:$\ \frac{(-2x+0.25y)}{(2y-0.25x)} $
The book talked something about solving for y (from the original equation): 
$\ \pm \sqrt{-x^2+0.25xy+16}$
then substituting it back into the implicitly derived equation and solving, but I am not sure that is going to work or else it is going to be very cumbersome. Anyone know how to solve this, perhaps in a simpler (possibly more correct) fashion? Thanks!
P.S. I spent 30mins figuring out the mathTex stuff so it looks pretty, it was the least I could do I suppose? Only for all you picky math geniuses who are going to help me! :)
 A: You have already implicitly derived the expression for $\frac{dy}{dx}$. You can set that equal to 1 and simplify to get $x = -y$. Substitute that in the original equation and you have a quadratic that can be easily solved. 
I got the solutions $x = \pm 8/3$ and $y = \mp 8/3$.
The book seems to be making a big deal out of a very simple question.
A: Since it's an equation of degree 2, the curve in question is a conic section. Looking at the coefficients, it's probably an ellipse. (The criterion for that is that the quadratic form should always be positive.)
The equation is symmetric in $x$ and $y$. In other words, a reflection at the line $y=x$ will map the ellipse to itself. This also means that the line $y=x$ is also a main axis.

Coincidentally, the question also asks for a tangent with exactly the same slope as the main axis. This can only happen at the points where the other main axis intersects the ellipse. In our case, the other axis is the line $y=-x$.
Hence, the points in question have the form $(x,-x)$ and can be obtained by solving the equation
$$ x^2 + 1/4·x^2 + x^2 = 16 $$
i.e. $x = \pm\sqrt{64/9} = \pm 8/3$.
