Find extreme values of ellipse

Question

I have the curve equation

$$ Ax^2 + By^2 + Cxy = 1 $$

which represents ellipse with center in (0, 0) and rotated some angle.

How can I find max X and Y values (not semi-axes) on this ellipse? Points are marked on the picture.

There x-es and y-es can not be moved left or right side, so I cannot find dx/dy and dy/dx.

Answer 1

Given $$ Ax^2 + Cxy + By^2 - 1 = 0 $$

solve for $x$ to get $$ x = {{ - Cy \pm \sqrt {C^2 y^2 - 4A\left( {By^2 - 1} \right)} } \over {2A}} $$

That means that fixing $y$ you get two solutions for $x$, until $y$ reaches the top (or lower) extreme where the two solutions are coincident, i.e. $$ \eqalign{ & C^2 y^2 - 4A\left( {By^2 - 1} \right) = 0 \Rightarrow \cr & \Rightarrow \left( {C^2 - 4AB} \right)y^2 = -4A \Rightarrow \cr & \Rightarrow y = \pm \sqrt {{{-4A} \over {C^2 - 4AB}}} \cr} $$

So $$ \left\{ \matrix{ y_{extr} = \pm \sqrt {{{4A} \over {-C^2 + 4AB}}} \hfill \cr x_{extr.y} = - {C \over {2A}}y_{extr} \hfill \cr} \right. $$

Then do the same in solving for $y$

Answer 2

I see two methods, both involving treating y as a function of x, to find the maximum and minimum values of y, then treating x as a function of y, to find the maximum and minimum values of x.

The simpler in calculation is to set the derivative equal to 0.

We can set the derivative of y with respect to x to 0. With $Ax^2+ By^2+ Cxy= 1$ we have $2Ax+ 2By\frac{dy}{dx}+ Cy+ Cx\frac{dy}{dx}= 0$. Setting $\frac{dy}{dx}= 0$, $2Ax+ Cy= 0$ so $y= -2\frac{A}{C}x$. Putting that into $Ax^2+ By^2+ Cxy= 1$ we have $Ax^2+ 4\frac{A^2B}{C^2}x^2- 2Ax^2= \left(4\frac{A^2B}{C^2}- A\right)x^2= 1$. $x= \pm\frac{1}{\sqrt{4\frac{A^2B}{C^2}- A}}$ gives the x values for which y takes it maximum and minimum values.

Do the same treating x as a function of y to find the x values for which x takes it maximum and minimum values.

Answer 3

Another way, use polar co-ordinates $x=r \cos t, y= r \sin t$ to get

$r^2(t)=\frac{1}{A \cos^2 t+ B\sin ^2 t+C \sin t \cos t}$

Let $u(t)=1/r^2(t)$ $\implies u(t)-\frac{A+B}{2}= \frac{A-B}{2} \cos 2t + \frac{c}{2}\sin 2t= \sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}~ \sin(2t+k)$ $\implies \frac{A+B}{2}-\sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}\le u(t) \le \frac{A+B}{2}+\sqrt{\frac{(A-B)^2}{4}+\frac{c^2}{4}}.$

From here one can get $r_{min}$ and $r_{max}$.