Distance point on ellipse to centre I'm trying to calculate the distance of a certain point of an ellipse to the centre of that ellipse:

The blue things are known: The lengths of the horizontal major radius and vertical minor radius and the angle of the red line and the x-axis. The red distance is the desired result. It is not given where on the ellipse the point is. It can be anywhere on the ellipse. Is this problem possible? If so, in which can this be solved? Thanks in advance!

After reading Kaj Hansen's comment and trying a bit this is what I did, it still won't work though.
In a triangle, $tan(\theta)=\frac{\text{opposite side}}{\text{adjecent side}}$. The slope of a line is $\frac{\Delta y}{\Delta x}$. Therefor the slope of the red line is $\tan(\theta)$; the formula of the line is $y=\tan(\theta)\cdot x$. 
The formula of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. When I put the two formulas together I get 
$$\frac{x^{2}}{a^{2}}+\frac{(\tan(\theta)\cdot x)^{2}}{b^{2}}=1$$
After a bit of rearranging:
$$x=\pm \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
$$$$
$$y=\tan(\theta)\cdot x$$
$$y=\pm \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} $$
Now with the help of Pythagoras' theorem $c=\sqrt{a^{2}+b^{2}}$ the red line should be 
$$\sqrt{\left ( \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}\right ) ^{2}+\left ( \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} \right )^{2}}$$
which can be simplified:
$$\sqrt{\frac{a^{2}\cdot b^{2} \cdot (\tan(\theta))^{2}+a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$
This, however, does not give the right answer. Let's try something:
$a=2$; $b=1$; $\theta=\frac{1}{2}\cdot \pi$ (The point is the point where the ellipse intersects with the minor radius)
$$\sqrt{\frac{2^{2}\cdot 1^{2} \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+2^{2}\cdot 1^{2}}{2^{2}\cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+1^{2}}}$$
$$\sqrt{\frac{4 \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+4}{4\cdot (\tan(\frac{1}{2} \pi))^{2}+1}}$$
But wait, $\tan(\frac{1}{2}\cdot \pi)$ is undefined. The formula cannot be filled in completely, which is a requirement. I need a formula that can be filled in for every value of $\theta$ on the domain $[0,\frac{1}{2}\cdot \pi]$

We know that the Parametric equation of an Ellipse - not centered, and not parallel to the Axises - is:
$$
x(\alpha) = R_x \cos(\alpha) \cos(\theta) - R_y \sin(\alpha) \sin(\theta) + C_x \\
y(\alpha) = R_x \cos(\alpha) \sin(\theta) + R_y \sin(\alpha) \cos(\theta) + C_y
$$
Where:
- $C_x$ is center X.
- $C_y$ is center Y.
- $R_x$ is the major radius.
- $R_y$ is the minor radius.
- $\alpha$ is  the parameter, which ranges from 0 to 2π radians.
- $\theta$ is the Ellipse rotation angle.
You can see this equation explanation here: What is the parametric equation of a rotated Ellipse (given the angle of rotation)
So all you need to do is enter your parameters to this equation to find the point (X, Y) and calculate the distance fron the Ellipse center to point (X, Y) on the Ellipse edge. 
Distance equation:
$$
distance = \sqrt{(x_1 - x_2) ^ 2 + (y_1 - y_2) ^ 2}
$$
