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help ellipse problem solving

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rcmel

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1. Foci (5,-3) and (-3,-2) and major axix is equal to twice to the minor axis?
2. center (1,3) and passing through (1,2) (5,2)?
 
there may be more efficient way but one can go step by step and construct points, then solve ellipse using simultaneous equations:

1. find center (interpolation):
x=[(-3)+(5)]/2=1
y=[(-2)+(-3)]/2=-2.5

center is (1,-5/2)

2. find line that goes through both foci (just a warm up exercise):
y=mx+b // line representation

m= slope
b = y intercept

m=(y2-y1)/(x2-x1)= -1/8

using known point on the line (one of foci such as x=5, y=-3) we can determine b:
y=mx+b
-3=(-1/8)*5+b
b= -19/8

so line is
y= -(x+19)/8

3. find point(s) on major axis. we know that this is along line that passes through both foci and distance from centre is twice the distance of foci from centre
suppose we look at foci point (5,-3) and we know centre is at (1, -5/2). then distance between them (along x and y axes) is
dx=5-1=4
dy=-3-(-5/2)=-1/2

therefore, point on major axis close to that foci is
(5,-3)+(4, -1/2) = (9, -7/2)
to verify result, we can check if this is on the line we just calculated
-7/2=-(9+19)/8
-7/2=-28/8
which is true

this is one of the points on ellipse, mark it for later use: (9, -7/2)

4. do the same for other foci, note dx and dy are same but displacement is now in opposite direction from the foci (hence negative sign):
(-3,-2) - (4, -1/2)= (-7, -3/2) // this should be the second point, verify it

5. find line that is normal to one going through foci (you had warmup, now do this on your own). we know that slope of perpendicular line is
m2 = -1/m1 = -1/(-1/8) = 8
and we know that this line also must go through center of the ellipse (so we have reference point and can calculate intercept)

6. find point that is on this line (minor axis line as found in step 5) and that is same distance from center as one of the foci.
the other way to deal with this is to just swap dx and dy:
dx=-1/2
dy=4

and add or subtract them from center (1,-5/2)

(1,-5/2) + (-1/2,4) = (1/2, 3/2) // this is another point on ellipse
(1, -5/2) - (-1/2, 4) = (3/2, -13/2) // this is another point on ellipse

7. so far we have found several points on ellipse. now you could go to wiki and read about ellipse and all the goodies like eccentricity etc or you could just write formula and solve it for known points (system of equations:

x^2/a^2 - y^2/b^2=1
 
continued....

so points we have found on ellipse are
(9, -7/2)
(-7, -3/2)
(1/2, 3/2)
(3/2, -13/2)

for example using first one we get
x=9, y= -7/2. plug those values into equation of ellipse and you get equation with two unknowns (a and b)
do this for all points and you got the system of equations (redundant but that's ok).
solve it any way you like (gaussian elimination, matrices, LU decomposition, adding/subtracting equations etc.)
 
continued....
question 2 is incomplete, so there are infinitely many solutions.
if we assume that same rule applies that major axis is twice the minor axis there is finite number of solutions
 
just have a one question in slope,. what is that -1/8?
thanks a lot and God Bless
 
Last edited:
rcmel,

1. Foci (5,-3) and (-3,-2) and major axix is equal to twice to the minor axis?

I can contruct the ellipse from the above information. The ellipse is rotated 7.13° clockwise, so it will have a xy term. The solution you were given does not have that term. Therefore, that solution is wrong. Before I show you the solution, I will let you work on it some more.

center (1,3) and passing through (1,2) (5,2)?

I have no idea what the above means.

Ratch
 
i stand corrected. :eek:

Ratchit is correct, here is equation for tilted one:
**broken link removed**
 
rcmel,

As you must know by now, an ellipse is the locus of points whose sum of distances from each focus point is a constant. The first thing you must do is calculate this constant, call it "s". Since we know the major axis is twice the minor axis, we can set up and solve for s in the following equation.

2(s-√65)+√65 =2*2(√((s/2)^2-(√65/2)^2)) ===>2*s-√65=2√(s^2-65)===>s=(5√65)/4

Now that you know what "s" is, you should be able to set up an equation for this particular ellipse.

Ratch
 
To the Ineffable All,

Seems like that are a lot of hits on this thread, so I will continue the problem. See the first attachment for setting up the ellipse equation and the second attachment for rationalizing it.

Ratch
 

Attachments

  • Ellipse_A.JPG
    Ellipse_A.JPG
    50 KB · Views: 263
  • Ellipse_B.JPG
    Ellipse_B.JPG
    82.1 KB · Views: 282
Hello there,


I havent gone over this thread in detial but I just wanted to mention that quite often the next step is to rotate the ellipse into standard position. Ratchit if you'd like to show that too that could help somewhat. The ellipse is rotated by the appropriate angle and offset by the appropriate x and y offset values.
 
MrAl,

It looks like a made a mistake in calculating "s". The correct value is (2/3)*√195. The attachments show the equations and the plots.

Ratch
 

Attachments

  • Ellipse_A.JPG
    Ellipse_A.JPG
    47.1 KB · Views: 305
  • Ellipse_B.JPG
    Ellipse_B.JPG
    38 KB · Views: 280
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