How to find the outermost points in an ellipse? If an ellipse is given in the form:
$$
A(x − h)^2+ B(x − h)(y − k) + C(y − k)^2 = 1
$$
(where A, B, C, h, and k are given)
What would be the simplest way of finding the outermost points, by which I mean the points both lowest and highest on the x-axis as well as the y-axis without graphing?
Edit: By lowest and highest points I mean the extremes, and the solution should be the four extreme points.
 A: Hints: 


*

*It can be hyperbola also. 

*The form shows it is capable of being displaced to origin by a shift (h,k) so that it becomes a central conic in form $ A_1 x^2 + B_1 x y + C_1 y^2 = 1 $.

*Next, find the tilt angle to align axes along x and y.

*By symmetry find maximum/minimum points along x- and y- axes.

*Shift back, rotate back.
A: To find the highest and lowest values of $y$ differentiate with respect to $x$ and set $y'=0$
$$2A(x-h)+B(y-k)=0$$ so that $(x-h)=-\frac B{2A}(y-k)$ and substituting into the original equation gives the quadratic $$\left(\frac {B^2}{4A}-\frac {B^2}{2A}+C\right)(y-k)^2=1$$ which can easily be solved for $(y-k)$ and hence for $y$ (the quadratic gives the two values you expect).
Then do the same with $x$, or just write down the formula using the symmetry of the expression.
A: Assume you've got a continuous, differentiable parametrization $(x(t), y(t))$ of the curve (you don't need to write it down explicitly, it suffices that we know that it exists). I'll use the Newton notation $\dot x = \mathrm dx/\mathrm dt$ and $\dot y = \mathrm dy/\mathrm dt$.
The extrema in $x$ are given by $\dot x=0$, and the extrema in $y$ are given by $\dot y=0$ (note that those are in general not the extrema of the ellipsis, that is, the extrema in the distance from the center). Note that if the ellipsis is non-degenerate (and the parametrization is sane), you never have both $\dot x=0$ and $\dot y=0$ for the same point. I'll assume a non-degenerate ellipse in the following.
Deriving the equation for $t$ gives
$$2A(x-h)\dot x + B(y-k)\dot x + B(x-h)\dot y + 2C(y-k)\dot y = 0.\tag{*}$$
Inserting $\dot x=0$ gives
$$(B(x-h) + 2C(y-k))\dot y = 0 \tag{**}$$
and therefore
$$(y-k) = -\frac{B}{2C}(x-h).$$
Inserting into the original equation you get
$$A(x-h)^2 - \frac{B^2}{2C}(x-h)^2 + \frac{B^2}{4C}(x-h)^2 = 1$$
or
$$x = h\pm\left(A - \frac{B^2}{4C}\right)^{-1/2}.$$
The corresponding values for $y$ are obtained by inserting this into (**) and solving for $y$.
The extrema for $y$ can be obtained analogously by setting $\dot y=0$ in (*) and doing the same procedure.
