areaconic sectionsgeometry
One can prove the ellipse area formula $A=\pi a b$ ($a$, $b$ the major and minor semi-axis) either by integration or by the stretched-circle argument. See for instance here: https://proofwiki.org/wiki/Area_of_Ellipse
Are there any other proofs of this formula?
Part 1: \begin{align} x(t) &= \cos(\alpha) \cos(t) - h\sin(\alpha) \sin(t) \\ y(t) &= \sin(\alpha) \cos(t) + h\cos(\alpha) \sin(t) \end{align}
I don't understand the question for part 2. Are the two point supposed to be on the ellipse? If so, then $h$ and $\alpha$ cannot be determined, because there can multiple ellipses at the origin containing the same two points. (Quick proof: draw a (non-circular) ellipse $E$ at the origin. Rotate it by 30 degrees to get another ellipse $E'$. Now look at $E \cap E'$. It will typically contain four points. If any two of these are called $A$ and $B$, then both $E$ and $E'$ are possible solutions to the "find $h$ and $\alpha$" problem, but the $\alpha$ values for $E$ and $E'$ differ by 30 degrees.
Post-comment remarks
The squared distance from $(x(t), y(t))$ to $(0,0)$ is \begin{align} d(t)^2 &= \left[\cos(\alpha) \cos(t) - h\sin(\alpha) \sin(t)\right]^2 + \left[\sin(\alpha) \cos(t) + h\cos(\alpha) \sin(t)\right]^2 \\ &= \cos^2(\alpha) \cos^2(t) + h^2\sin^2(\alpha) \sin^2(t) - 2h\cos(\alpha) \cos(t)\sin(\alpha) \sin(t) + \sin^2(\alpha) \cos^2(t) + h^2\cos^2(\alpha) \sin^2(t) + 2h\cos(\alpha) \cos(t)\sin(\alpha) \sin(t) \\ &= (\cos^2(\alpha) + \sin^2(\alpha)) \cos^2(t) + h^2(\sin^2(\alpha) + \sin^2(t)) \sin^2(t) \\ &= \cos^2(t) + h^2 \sin^2(t) \end{align} This varies from $1$ (at $t = 0$) to $h^2$ (at $t = \frac{\pi}{2}$). Hence one (semi-) axis has length $1$ and the other has length $h$. If $h < 1$, then the major (semi-)axis will have length $1$.
Note that at $t = 0$, we have $$ \begin{align} x(0) &= \cos(\alpha) \cos(0) - h\sin(\alpha) \sin(0) &= \cos(\alpha) \ y(0) &= \sin(\alpha) \cos(0) + h\cos(\alpha) \sin(0) &= \sin(\alpha) \end{align} which is an angle $\alpha$ from the positive $x$-axis, as required.
It's possible that you still don't believe me, so here's some Matlab code:
function y = ellipseTest()
hList = [0.1, 0.3, 0.7]; % three different eccentricities
alphaList = [20, 50, 130] * pi/180; % three different major-axis angles
t = linspace(0, 2*pi, 100); % sample values from 0 to 2pi
colorList = ['r', 'g', 'b'];
figure(gcf);
clf;
hold on
for i = 1:3
h = hList(i); alpha = alphaList(i); color = colorList(i);
x = cos(alpha) * cos(t) - h * sin(alpha) * sin(t);
y = sin(alpha) * cos(t) + h * cos(alpha) * sin(t);
plot(x, y, color)
end
hold off;
set(gca, 'DataAspectRatio', [1 1 1]);
figure(gcf);
and the resulting figure:
I find these pretty compelling.
Note that each pair of ellipses intersects in two points in the first quadrant, so if you picked one of these sets-of-two-points for your part 2 question, you'd find two different ellipses that fit them.
Recall that any ellipse can be obtained by affinity from a circle and that the affinity preserves the ratio between the areas.
Thus we can consider the division into sectors of equal area for a circle and then the corresponding affine transformation to the ellipse.
Notably. as an example, let consider the affinity by $X=ax$ and $Y=by$ such that
$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$
with reference to the first quadrant, the equations for the lines dividing the circle into $n$ sectors of equal area $A=\frac{\pi}{4n}$ are
$$y=(\tan \theta_k)\cdot x, \quad \theta_k=\frac{k\pi}{2n},\quad k=1,...,n-1$$
the equations for the lines dividing the ellipse into $n$ sectors of equal area $A=\frac{\pi ab}{4n}$ are
$$Y=\frac b a (\tan \theta_k)\cdot X, \quad \theta_k=\frac{k\pi}{2n},\quad k=1,...,n-1$$
Best Answer
Another way could be via Cavalieri's principle. Suppose we have an ellipse with semi-axes $a$ and $b$ and construct a circle with radius $\sqrt{ab}$. Draw a pair of lines $FF'$, $EE'$ tangent to the circle at the endpoints of a diameter (see figure below) and rotate the ellipse so that it has those lines as tangents. If we show that any chord $CD$ of the circle, parallel to $FF'$, has the same length as the corresponding chord $C'D'$ of the ellipse, then circle and ellipse have the same area $\pi ab$.
Let's consider first of all the diameter $AB$ of the circle parallel to $FF'$: the corresponding diameter $A'B'$ of the ellipse forms with $E'F'$ a pair of conjugate diameters. It is well known that the parallelogram formed by semi-diameters $O'B'$ and $O'F'$ has area $ab$: as the height of such parallelogram with respect to base $O'B'$ is $\sqrt{ab}$ we deduce then $O'B'=\sqrt{ab}$. It follows that $AB=A'B'$.
Let's now prove the analogous equality for any other couple of corresponding chords $CD$ and $C'D'$. The equation of conjugate diameters states that $$ {O'M'^2\over O'F'^2}+{C'M'^2\over O'A'^2}=1, $$ but on the other hand we also have: $$ {O'M'\over O'F'}={OM\over OF}={OM\over \sqrt{ab}}. $$ Inserting this into the previous equality, and taking into account that $O'A'=\sqrt{ab}$ we get: $$ C'M'=\sqrt{ab-OM^2}=CM, $$ where we also used Pythagoras' theorem applied to triangle $OMC$. This completes the proof, because $M'$ is the midpoint of $C'D'$.