Line Intersection with an Ellipse

[kylbrrws]

Hello,

I am a software developer and I need to plot a point on an Ellipse. Unfortunately I am not that good at Math, I've done a lot of research ad still don't quite understand the formulas.

Say my ellipse is centered at: x,y (3,4) and it is a=55, b=125. My line begins at the ellipse center point and goes 30 degrees from the center point until it hits the ellipse path, I need to find the point that it hits the path at.

Can someone show the work that needs to be done in addition to the formula -because this is what trips me up -in comp programming math is expressed very differently.

One additional question - does this formula work for an ellipse centered at 0,0?

-Kyle

 

[Deleted member 4993]

Hello,

I am a software developer and I need to plot a point on an Ellipse. Unfortunately I am not that good at Math, I've done a lot of research ad still don't quite understand the formulas.

Say my ellipse is centered at: x,y (3,4) and it is a=55, b=125. My line begins at the ellipse center point and goes 30 degrees from the center point until it hits the ellipse path, I need to find the point that it hits the path at.

Can someone show the work that needs to be done in addition to the formula -because this is what trips me up -in comp programming math is expressed very differently.

One additional question - does this formula work for an ellipse centered at 0,0?

-Kyle

The equation of an ellipse whose axes are parallel to x & y axes and centered at (h,k) with semi-axes length of a & b:

$\displaystyle \dfrac{(x-h)^2}{a^2} \ + \ \dfrac{(y-k)^2}{b^2} \ = \ 1$

equation of line going through (h,k) with a slope = m = tan(Θ):

y - k = m * (x - h)

Now calculate the common 'x' and 'y' from the above two equations.

 

[kylbrrws]

So this is what I am seeing from your reply..

m = tan(30)
ellipse = (x-3)^2 / 55^2 + (y-4)^2 / 125^2


I don't understand how to get the x & y

 

[Deleted member 4993]

So this is what I am seeing from your reply..

m = tan(30)
ellipse = (x-3)^2 / 55^2 + (y-4)^2 / 125^2


I don't understand how to get the x & y

There two equations -

(x-3)^2 / 55^2 + (y-4)^2 / 125^2 = 1

and

y-4 = (1/√3)*(x-3)

and you have two unknowns (x & y)

let

u = x-3 and.................................(1)

v = y-4 .................................(2)

then we have

u^2 / 55^2 + v^2 / 125^2 = 1 .................................(3)

and

v = (1/√3)*u .........................................................(4)

using (4) in (3), we get

u^2 / 55^2 + u^2 /(3* 125^2) = 1 ..........................(5)

Now solve for 'u' from (5) - use that value of 'u' in (4) and calculate 'v' - then calculate 'x' and 'y' using (1) and (2)

 

[kylbrrws]

There two equations -

(x-3)^2 / 55^2 + (y-4)^2 / 125^2 = 1

and

y-4 = (1/√3)*(x-3)

and you have two unknowns (x & y)

let

u = x-3 and.................................(1)

v = y-4 .................................(2)

then we have

u^2 / 55^2 + v^2 / 125^2 = 1 .................................(3)

and

v = (1/√3)*u .........................................................(4)

using (4) in (3), we get

u^2 / 55^2 + u^2 /(3* 125^2) = 1 ..........................(5)

Now solve for 'u' from (5) - use that value of 'u' in (4) and calculate 'v' - then calculate 'x' and 'y' using (1) and (2)

Can you show me how to get x and y on the other side of the equation..
as in...

x=?
y=?

As I said before - I am not doing this for school or something - so I have really no Math experience on this level ~ when you say "solve for 'u' ", I don't know what to do.

 

[HallsofIvy]

One thing that follows from the "circle" definitions of sine and cosine is that if (x, y) is a point on the circle with center $\displaystyle (x_0, y_0)$ and radius r has parametric equations $\displaystyle x= x_0+ r cos(\theta)$ and $\displaystyle y= y_0+ r sin(\theta)$. Taking $\displaystyle \theta$ from 0 to $\displaystyle 2\pi$

It shouldn't be difficult to see that an ellipse with axes parallel to the x-axis of length L and parallel to the y-axis of length H is given by $\displaystyle x= x_0+ L cos(\theta)$ and $\displaystyle y= y_0+ H sin(\theta)$.

A line that goes through $\displaystyle (x_0, y_0)$ and makes angle $\displaystyle \phi$ with the positive x-axis is given by $\displaystyle x= x_0+ t cos(\phi)$ and $\displaystyle y= y_0+ t sin(\phi)$ with parameter t.

The points where the line intersects the ellipse are given by $\displaystyle x= x_0+ L cos(\theta)= x_0+ t cos(\phi)$ and $\displaystyle y= y_0+ Hcos(\theta)= y_0+ t cos(\phi)$. Here, $\displaystyle x_0$, $\displaystyle y_0$, L, H, and $\displaystyle \phi$ are given constants (in your problem $\displaystyle x_0= 3$, $\displaystyle y_0= 4$, L= 55, H= 125, and $\displaystyle \phi= 30$ degrees. That gives two equations to solve for t and $\displaystyle \theta$.

I think you will find it much easier to plot points in computer graphics using "parametric equations" where x and y are given in terms of some third parameter.