Distance of point on ellipse from origin

Question

I need to determine distance to a point on an ellipse.

center point is 0,0 (Cartesian coordinates).

I have a,b. for example, b (y radius) is .66, a (x radius) is .87

I have angle from positive X axis (0 quadrant), 45 or 30 degrees for example.

wouldn't the distance (ds) to the point simply be:

d (difference) = abs (a-b). example: abs(.66-.87) = .21

rmin (minimum radius) = smallest of a and b.

r (ratio) = angle/90. examples: .5 for 45/90, or .33 for 30/90.

formula: ds = rmin + (d * r) or distance = minimum radius plus (difference * ratio)

if 45 degrees = .66 + (.5 * .21) = .765. similarly, .73 for 30 degrees

(adjustments would have to be made for other quadrants). Thanks!

Answer 1

You can write the ellipse as such: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \iff$

$$\frac{(rcosθ)^2}{a^2}+\frac{(rsinθ)^2}{b^2}=1$$

Since you know $a,b$ and $θ$ of your point, you can solve for $r$ and that is the distance of the point from the origin.

(For better formatting you may want to have look here: MathJax basic tutorial and quick reference)

Answer 2

Take the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ use a cartesian to polar conversion $x=r \cos\theta$ and $y=r \sin \theta$ and solve for $r$

$$ \frac{ r^2 \cos^2 \theta}{a^2} + \frac{ r^2 \sin^2 \theta}{b^2} = 1 $$

$$ \boxed{ r = \frac{1}{\sqrt{ \frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2} } } } $$

Plug in your angle $\theta=30°$ and the semi-radii $a=0.87$ and $b=0.66$ to get for example $r =\frac{1}{\sqrt{0.990883868410622+0.573921028466483 }} = 0.799410598404111 $