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 $[0,\frac{1}{2}\cdot \pi]$

I'm assuming that the centre is located on the origin.
Let us call the point where the red line meets the ellipse P. Let P have coordinates $(x_1, y_1) $. Notice that $tan\theta = y_1/x_1$. Let the horizontal semi-axis be $a$, meaning the ellipse is $2a$ wide. Similarly, let the vertical semi-axis be $b$.
Let the known blue angle be $\theta$.
Equation of an ellipse(centered at origin) =
$x^2/a^2 + y^2/b^2 = 1$
Since $(x_1,y_1)$ satisfies the equation, the equation becomes:
$x_1^2/a^2 + y_1^2/b^2 = 1$
But $tan\theta = y_1/x_1$ so $y_1^2 = x_1^2tan^2\theta$. Substituting this:
$x_1^2/a^2 + x_1^2tan^2\theta/b^2 = 1$
$x_1^2(1/a^2 + tan^2\theta/b^2) = 1$
$x_1^2(\frac{a^2tan^2\theta+b^2}{a^2b^2}) = 1$
$x_1^2= \frac{a^2b^2}{a^2tan^2\theta+b^2}$
Finding the value of $y_1^2$, we get:
$y_1^2 = \frac{a^2b^2}{a^2tan^2\theta+b^2}\times tan^2\theta$
$y_1^2 = \frac{a^2b^2tan^2\theta}{a^2tan^2\theta+b^2}$
Since we have assumed the centre to be the origin, the distance becomes: 
$\sqrt{x_1^2+y_1^2}$
$\sqrt{\frac{a^2b^2}{a^2tan^2\theta+b^2}+\frac{a^2b^2tan^2\theta}{a^2tan^2\theta+b^2}}$
Since $1+tan^2\theta = sec^2\theta$, we have:
$\sqrt{\frac{a^2b^2sec^2\theta}{a^2tan^2\theta+b^2}}$
Addressing your problem, note that both tan and sec don't have continuous domains, however, both of them have cos as the denominator in one or the other representation. So, we can multiply both the numerator and denominator by $cos\theta$ to get rid of the discontinuous domain.
$\sqrt{\frac{a^2b^2sec^2\theta}{a^2tan^2\theta+b^2}}\times\sqrt{\frac{cos^2\theta}{cos^2\theta}}$
$\sqrt{\frac{a^2b^2}{a^2sin^2\theta+b^2cos^2\theta}}$
$\frac{ab}{\sqrt{a^2sin^2\theta+b^2cos^2\theta}}$
You could rationalize the denominator if you want. This formula works for $\theta = \pi/2$
Note: Even though we have assumed the ellipse to be centred at origin, the formula still works as any other ellipse with same a,b would be congruent to the discussed one.
