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 $[0,\frac{1}{2}\cdot \pi]$
If $a$ is the length of the horizontal axis and $b$ the vertical axis, then the equation for the ellipse is $(x/a)^2 + (y/b)^2 = 1$.  

From here, we can also get an equation for the red line since its slope is $\tan(\theta)$ and $y$-intercept is $0$.

To find the point where the line intersects the ellipse, simply take the equation of the line, $y = mx$, and plug it into the equation for the ellipse, $(x/a)^2 + (y/b)^2 = 1$, and solve for $x$ (note that this will give two possible values).  Then, you can plug the appropriate $x$-value back into the ellipse formula to get the corresponding $y$-coordinate.

Now that you have the point of intersection, simply use the Pythagorean theorem to obtain the length of the red segment.
Convert the equation of ellipse from rectangular to polar coordinates.

$ (x/a)^2 + (y/b)^2 = 1 $ , $ x = r cos(\theta), y = r sin(\theta)$.

$ (cos(\theta)/a )^2+ (sin(\theta)/b)^2 =1/r^2 $.

$(a,b,\theta )$ are given. Find out 1/$r^2$ and $r$.