With the center at $(0,0)$ and $\psi=0$, the equation of the ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{.}$$
So $$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$$ and therefore $$\frac{dy}{dx}= -{\frac{xb^2}{ya^2}}\text{.}$$
The slope of the normal line is the negative reciprocal of this, so $$\tan(\phi)=\frac{ya^2}{xb^2}\text{.}$$
Meanwhile, $$\tan(\theta)=\frac{y}{x}\text{.}$$ So, eliminating $\frac{y}{x}$ from these two equations and clearing denominators, the relationship between $\phi$ and $\theta$ is: $$b^2\tan(\phi)=a^2\tan(\theta)$$
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
$$
\begin{align}
x(0) &= \cos(\alpha) \cos(0) - h\sin(\alpha) \sin(0) &= \cos(\alpha) \
y(0) &= \sin(\alpha) \cos(0) + h\cos(\alpha) \sin(0) &= \sin(\alpha)
\end{align}
which is an angle $\alpha$ from the positive $x$-axis, as required.
It's possible that you still don't believe me, so here's some Matlab code:
function y = ellipseTest()
hList = [0.1, 0.3, 0.7]; % three different eccentricities
alphaList = [20, 50, 130] * pi/180; % three different major-axis angles
t = linspace(0, 2*pi, 100); % sample values from 0 to 2pi
colorList = ['r', 'g', 'b'];
figure(gcf);
clf;
hold on
for i = 1:3
h = hList(i); alpha = alphaList(i); color = colorList(i);
x = cos(alpha) * cos(t) - h * sin(alpha) * sin(t);
y = sin(alpha) * cos(t) + h * cos(alpha) * sin(t);
plot(x, y, color)
end
hold off;
set(gca, 'DataAspectRatio', [1 1 1]);
figure(gcf);
and the resulting figure:
I find these pretty compelling.
Note that each pair of ellipses intersects in two points in the first quadrant, so if you picked one of these sets-of-two-points for your part 2 question, you'd find two different ellipses that fit them.
Best Answer
Here's a way. Write $$ f(x, y) = \pmatrix{x & y & 1 } \pmatrix{a & b/2 & d/2 \\ b/2 & c & f/2 \\ d/2 & f/2 & g} \pmatrix{x\\y\\1}, $$ which I'll denote as $h^t M h$, where $M$ is the middle matrix, and $h$ is the vector $\pmatrix{x\\y\\1}$.
You'd like to find a matrix $T$ with the property that when $u^2 + v^2 = 1$, then $$ s = T\pmatrix{u\\v\\1} $$ has the property that $s^t M s = 0$. (I'm going to use a $3 \times 3$ matrix for $T$, but the last row will be $\pmatrix{0& 0&1}$, so you can chop it off to get your $T$ matrix, OK? )
Composing those two things, and letting w be the vector $\pmatrix{u\\v\\1}$ you want to have \begin{align} 0 &= s^t M s \\ 0 &= (Tw)^t M (Tw) \\ 0 &= w^t T^t M Tw \\ 0 &= w^(T^t M T)w \end{align} for all vectors $w$ with $u^2 + v^2 = 1$. So letting $K = T^t M T$, we want to have $$ \pmatrix{u & v & 1} K \pmatrix{u\\v\\1} = 0, $$ for all $u,v$ with $u^2 + v^2 = 1$. If we simply pick $K$ to be $\pmatrix{1&0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1}$, we get \begin{align} \pmatrix{u & v & 1} K\pmatrix{u\\v\\1} &= \pmatrix{u & v & 1} \pmatrix{1&0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1} \pmatrix{u\\v\\1}\\ &= u^2 + v^2 - 1 \\ &= 0. \end{align}
So now we know what we need: a matrix $T$ with the property that $T^t M T$ is the diagonal matrix with $1,1,-1$ on the diagonal.
To find this, we simply find the eigenvectors of $M$ and put them into the columns of $T$ and compute $T^t M T$; if the result has the $-1$ in the wrong spot, swap the vectors in $T$ to move that column to the last column. (BTW, you may also need to multiply each column by some constant to make the product have $1$s and $-1$ on the diagonal rahter than some other numbers).
Perhaps "find the eigenvectors of $M$" isn't something you know how to do -- you didn't really say how much math you knew -- but for this particular problem, it's not too tough. My best guess, off the top of my head, is that two of them look something like $$ \pmatrix {\cos \theta\\ \sin \theta \\ 0}, \pmatrix {-\sin \theta\\ \cos \theta \\ 0}, $$ where $\theta$ is the angle of rotation of your ellipse, and the third (with the negative eigenvalue) is the cross-product of these two. [I'm sure that guess is at least slightly wrong, but that's the gist.]