Solve for ellipse given axis-parallel tangent line and another tangent line See illustration I know point $P_1 = (x_1,y_1)$ which sits on a line $L_1$. $P_1$ will lie on the semi-minor axis. $L_1$ is parallel to the semi-major axis. Have another line $L_2$ which is tangent to the ellipse elsewhere in the first or fourth quadrants. I know a point $P_2 = (x_2,y_2)$ on this line which should also lie on the ellipse.
$P_1$, $P_2$, $L_1$, $L_2$ are all known. My problem is how do I find $a$ and $b$, the length of the semi-major and semi-minor axes? My intuition tells me my problem is constrained enough to be solved to a single ellipse.
I started with the Cartestian equation for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, did some implicit differentiation to get $\frac{dy}{dx}=\frac{-xb^2}{ya^2}$. I know the slope of the tangent line $L_2$, so thought this would be the best line of attack but have been staring at this ages now and don't know how to proceed. What I'm looking to get is two linear equations with $a$ and $b$ in them, right?
I went down the path of rearranging the derivative equation. I've substituted $m_2$ for $\frac{dy}{dx}$; the slope of $L_2$:
$$a = \sqrt{\frac{-x_2b^2}{y_2m_2}}$$
but don't see how this helps me because it's not a linear equation. How can I solve for my ellipse?
EDIT: I used the equation of an ellipse centred at the origin above, but this was a mistake. I'm happy to rotate/translate lines to make the calculations simpler and then transform back e.g. so that $x_1 = 0$.
 A: Since you do not actually have the ellipse's center given, you need a more general form for the ellipse's equation.
Here is a geometric hint; I leave it to you to translate this to algebra.
You know the minor symmetry axis of the ellipse:
It is the line orthogonal to $L_1$ passing through $P_1$.
Now all you need to determine the center $O$ is another non-parallel line
known to pass through $O$.
Let $M$ be the midpoint between $P_1$ and $P_2$.
Let $Q$ be the intersection point of $L_1$ and $L_2$.
Then the straight line $MQ$ passes through $O$.

Why? Because that would be the case if the ellipse were a circle.
Deforming a circle into an arbitrary ellipse can be done with an
invertible affine transformation.
Such transformations preserve line straightness, tangency and midpoint
relationships.
Therefore $MQ$ still passes through $O$ after the transformation.
A: Assume that $P_1=(0,b)$ and $P_2=(a\cos\theta, b\sin\theta)=(h,k)$, both lying on the ellipse $\frac {x^2}{a^2}+\frac{y^2}{b^2}=1$, and that $L_1: y=b$, and $L_2: x+By+C=0\tag{1}$. 
Quantities $b,h,k,B,C$ are known. 
We only need to find the value of $a$ to determine the ellipse.
First we can determine $\theta$ as equating $k=b\sin\theta$, and consequently determine a using $h=a\cos\theta$. 
By differentiating and using standard formulas, we find that the equation of the tangent at $P$ is
$$\frac x{a\sin\theta}+\frac y{b\cos\theta}-(\tan\theta+\cot\theta)=0$$
which can be written as
$$x+\left(\frac ab\tan\theta\right)y-a\sin\theta(\tan\theta+\cot\theta)=0\tag{2} $$
Using values of $a,\theta$ as computed, coefficients in equations $(1),(2)$ should turn out to be equal.
