Ellipse in Cartesian and Polar coordinates: x=2*cos(theta), y=1*sin(theta)

[ChazAshley]

Hello there.
Let's say we have a turret at the center of an ellipse. And when this turret turns, we want to get a point on the edge of the ellipse which the turret points on.

If we have kinda fat ellipse with radius = 2 along X axis, and radius = 1 along Y axis, we can get each point of ellipse like this

x=2*cos(theta)
y=1*sin(theta)

And when I calculate a point by equations above, it doesn't return the right point. I kinda understand, that by multiplying cos(theta) by 2 we just stretch a circle, and all points stretch as well.

When I asked smart people about this, they gave me the soultion - use ellipse equation in polar coordinates. The equation returns radius, so by multiplying cos(theta) and sin(theta) by that radius i'm able to get the right point.
Here is link to some animation that shows the difference in the point calcuated with and without using polar equation of ellipse.
https://www.desmos.com/calculator/cs9u28uxg0

The purple point is the point calculated with using polar equation of ellipse.

Could someone explain me why does this work like this, please?

 

[HallsofIvy]

Hello there.
Let's say we have a turret at the center of an ellipse. And when this turret turns, we want to get a point on the edge of the ellipse which the turret points on.

If we have kinda fat ellipse with radius = 2 along X axis, and radius = 1 along Y axis, we can get each point of ellipse like this

x=2*cos(theta)
y=1*sin(theta)

And when I calculate a point by equations above, it doesn't return the right point.

What do you mean by this? Strictly speaking these are "parametric equations" not polar equations but they certainly will give the point on the ellipse at angle theta with the positive x axis

I kinda understand, that by multiplying cos(theta) by 2 we just stretch a circle, and all points stretch as well.

When I asked smart people about this, they gave me the soultion - use ellipse equation in polar coordinates. The equation returns radius, so by multiplying cos(theta) and sin(theta) by that radius i'm able to get the right point.
Here is link to some animation that shows the difference in the point calcuated with and without using polar equation of ellipse.
https://www.desmos.com/calculator/cs9u28uxg0

The purple point is the point calculated with using polar equation of ellipse.

Could someone explain me why does this work like this, please?

 

[ChazAshley]

I mean, if we have an ellipse with big radius = 4, and little radius = 2, and then we draw a line that has 45 degrees angle, then i want to get the point on the ellipse where the line intersects the ellipse.
With parametric equation we will have
x=4*cos(pi/4) which is 2.82
y=2*sin(pi/4) which is 1.41
And this point isn't the point of intersection.

But if we will use the equation of ellipse in Polar coordinates, we can plug in those two radiuses and the angle, and get the length of the segment of the line from the center of coordinates to the point of intersection. And then we can use this length as radius in those two parametric equations to get the right point of intersection.

There is a link to the desmos calculator in my first message, so you can watch the dependency between the "real" and "wrong" points.