Find (x,y) of a Point on an Ellipse Given Foci and Radii Info

[Crusty]

expressing points on an ellipse given the (x,y) of the foci and the sum of the radii

Given the (x,y) of the 2 foci of an ellipse,
and the sum of the radii from the foci,

Is there an equation that will find the (x,y) of a point on the ellipse at a specified angle?

Is there an equation to find the (x,y) given one radius' length from a specified focus?

Thank you.

 

[HallsofIvy]

The total distance from a point on an ellipse to the two foci is always the same. I assume that by "sum of the radii from the foci" you mean that number. Imagine an ellipse in "standard position" (center at (0,0), major and minor axes on the x and y axes) with vertices at (a,0),(-a,0),(0,b),(0,-b) and foci at (c,0), (-c,0). Then the line from the point (0,b) to (c,0) is the hypotenuse of a right triangle with legs of length b and c. The distance from (0,b) to (c,0) is $\sqrt{b^2+ c^2}$ and so the total distance from (0,b) to both foci is $2\sqrt{b^2+ c^2}$.

The disance from the focus (c,0) to the point (a, 0) is a- c, of course, and the distance from the focus (-c,0) to the point (a,0) is a-(-c)= a+ c. The total distance is (a- c)+ (a+ c)= 2a.

Since those two total distances must be the same [itex]2a= 2\sqrt{b^2+ c^2} or a²= b²+ c².

Thus if you are given the (x,y) coordinates of the foci, c is half the distance between them. If you are also given the total distance from each point on the ellipse to the foci, a is half that distance and b can be calculated from formula above.

You will still need to account for the angle the major axis makes with the x-axis but that is just the arctan of the slope of the line through the two foci.

 

[Crusty]

Thanks
That's good. I think I was just before that point of trying those known points on the ellipse with right triangles. Sounds correct. Is there a math word for sum of radii in an ellipse?

The method you show should work and I may use it in the end. But I was checking another method too. I'll finish checking this one myself later when I have more time, but in the mean time if you catch this would you mind correcting it?

2nd radius = ( sqrt( (distance between foci)^2 + ( (sum of radii) - (2nd radius) )^2 )

um, how do you get the 2 instancies of (2nd radius) to the same side of the equation?

2nd radius = ( (distance between foci) / (sum of radii) + (sum of radii)
) / 2

arg those two (sum of radii) don't look right...

It was based on a right triangle with both foci as points and the right angle at the 1st focus, rather than at the mid point. So if it works for that, I'd still need to check if it would work for other angles...