# Expressing points on an ellipse given the (x,y) of the foci and the sum of the radi

1. Mar 17, 2006

### 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.

Last edited: Mar 17, 2006
2. Mar 17, 2006

### 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.

3. Mar 17, 2006

### 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?