Advanced Ellipse Constructions

Steven Dutch, Professor Emeritus, Natural and Applied Sciences,University of Wisconsin - Green Bay

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Finding the Axes of an Ellipse

Given an ellipse with unknown axes: Draw two parallel chords to the ellipse and bisect them (purple). Draw a line (blue) through the bisection points and extend it until it intersects the ellipse on both ends. Bisect the portion of the line within the ellipse. The center point of the line is the center of the ellipse. If the ellipse is somewhat rough, do this construction several times to get the best approximation.

Construct a circle centered on C with radius sufficient to intersect the ellipse in four places (blue). The chords of the circle that connect the intersection points are parallel to the axes of the ellipse. The four chords should form a rectangle. Alternatively, draw lines between opposite intersection points and bisect the resulting angles to find the axes.

Finding an Ellipse Given Five Points

The Mathematical Solution

The general formula for a conic section is Ax² + Bxy + Cy² + Dx + Ey + F = 0. We eliminate F usually by dividing all the terms in the equation by F to make the final term 1.

Given five points (x₁,y₁); (x₂,y₂); (x₃,y₃); (x₄,y₄); (x₅,y₅),  we can plug the coordinates in to the general formula to get five equations in five unknowns (A',B',C',D',E', where A' = A/f, etc.):

A'x₁² + B'x₁y₁ + C'y₁² + D'x₁ + E'y₁ + 1 = 0
A'x₂² + B'x₂y₂ + C'y₂² + D'x₂ + E'y₂ + 1 = 0
A'x₃² + B'x₃y₃ + C'y₃² + D'x₃ + E'y₃ + 1 = 0
A'x₄² + B'x₄y₄ + C'y₄² + D'x₄ + E'y₄ + 1 = 0
A'x₅² + B'x₅y₅ + C'y₅² + D'x₅ + E'y₅ + 1 = 0

We can now set up a family of determinants:

        |x₁²  x₁y₁  y₁²  x₁  y₁|                             |1  x₁y₁  y₁²  x₁  y₁|
        |x₂²  x₂y₂  y₂²  x₂  y₂|                             |1  x₂y₂  y₂²  x₂  y₂|
D =  |x₃²  x₃y₃  y₃²  x₃  y₃|                   D_A =  |1  x₃y₃  y₃²  x₃  y₃|
        |x₄²  x₄y₄  y₄²  x₄  y₄|                             |1  x₄y₄  y₄²  x₄  y₄|
        |x₅²  x₅y₅  y₅²  x₅  y₅|                             |1  x₅y₅  y₅²  x₅  y₅|, etc.

We evaluate these using very tedious but very standard algebra, then we find A = D_A/d, etc. Lots of spreadsheets will do determinants, or if you're coding this in a program, once you've written the code for one determinant, simple editing will rapidly generate the rest.

Once we have A', B', C', D', and E', what next?

We can solve for the axes, orientation and center of the ellipse but we need to develop the general equation of an ellipse in terms of those parameters first.

General Equation of an Ellipse

Start with an ellipse in standard Cartesian coordinates, then rotate it and translate it to get the general equation:

x²/a² + y²/b² = 1 or, clearing fractions

b²x² + a²y² = a²b²

Rotate by angle w:

x' = xcosw - ysinw    y'= xsinw + ycosw

Theellipse becomes

b²(xcosw - ysinw)² + a²(xsinw + ycosw)² = a²b²

b²(x²cos²w -2xycoswsinw + y²sin²w) + a²(x²sin²w + 2xysinwcosw + y²cos²w) = a²b²

x²[b²cos²w + a²sin²w] + 2(a² - b²)xysinwcosw + y²[b²sin²w + a²cos²w] = a²b²

Finally translate the ellipse to center at (h,k); x becomes x-h and y becomes y-k.

(x² - 2hx +h²)[b²cos²w + a²sin²w]  + 2(a²- b²)(x-h)(y-k)sinwcosw + (y² - 2yk +k²)[b²sin²w + a²cos²w] = a²b²

Becomes:

• x²[b²cos²w + a²sin²w]
• + 2(a² - b²)(xy)sinwcosw
• + y²[b²sin²w + a²cos²w]
• - 2hx[b²cos²w + a²sin²w] + 2(a² - b²)(-kx)sinwcosw
• - 2yk[b²sin²w + a²cos²w] + 2(a² - b²)(-hy)sinwcosw
• + h²[b²cos²w + a²sin²w] + 2(a² - b²)(hk)sinwcosw + k²[b²sin²w + a²cos²w]
• = a²b²

In standard form, Ax² + Bxy + Cy² + Dx + Ey + F = 0

• A = [b²cos²w + a²sin²w]
• B = 2(a² - b²)sinwcosw
• C = [b²sin²w + a²cos²w]
• D = - 2h[b²cos²w + a²sin²w] - 2k(a² - b²)sinwcosw = -2hA - kB
• E = - 2k[b²sin²w + a²cos²w] - 2h(a² - b²)sinwcosw = -2kC - hB
• F = h²[b²cos²w + a²sin²w] + 2(a² - b²)(hk)sinwcosw + k²[b²sin²w + a²cos²w] - a²b²
  = h²A + (hk)B + k²C - a²b²

When we actually solve for our ellipse, we will have five sets of (x,y) coordinates and solve for the coefficients. We can then use the coefficients to find the parameters a, b, h, k, and w. But what will we do about the F term? We normalize the equations by dividing by F and reducing the constant term to 1. We will actually solve for A',B', etc. A = FA', etc.

Solving for Rotation Angle w

In A, B and C, the unknowns are a, b and w.

We can write B = (a² - b²)sin2w

Now cos2w = cos²w - sin²w = 2cos²w - 1 = 1 - 2sin²w. Hence

• cos²w  = (1+ cos2w )/2
• sin²w = (1 - cos2w)/2

Thus A = [b²cos²w + a²sin²w] = [b²(1+ cos2w)/2 + a²(1 - cos2w)/2] = (a² + b²)/2 - (a² - b²)(cos2w)/2

Also C = [b²sin²w + a²cos²w] =  [b²(1 - cos2w)/2 + a²(1+ cos2w )/2] = (a² + b²)/2 + (a² - b²)(cos2w)/2

Hence C - A = (a² - b²)cos2w. Note also that C + A = a² + b²

We can combine to get B/(C-A) = (a² - b²)sin2w/(a² - b²)cos2w = sin2w/cos2w = tan2w

Hence, tan2w = B/(C-A)

Since B/(C-A), we have FB'/(F(C'-A')) = B'/(C'-A'), so tan2w = B'/(C' - A')

It will be useful to have sin2w and cos2w in terms of A, B, and C.

• Sin2w = tan2w/�ˆš(1 + tan²2w) and cos2w = 1/�ˆš(1 + tan²2w).
• Tan2w = B/(C-A) so Sin2w = B/(C-A) /�ˆš(1 + B²/(C-A)²). Clearing fractions, we get
• Sin2w = B/�ˆš((C-A)² + B²).
• Cos2w = 1/�ˆš(1 + tan²2w) = 1/�ˆš(1 + B²/(C-A)²) = (C-A)/�ˆš((C-A)² + B²)

Also, C + A = (a² + b²) and B = (a² - b²)sin2w. We now know w, so sin2w is known.

Solving for Axes a, b

• We have C + A = (a² + b²), hence b² = C + A - a², also  
• B = (a² - b²)sin2w = (a² - C - A + a²)sin2w = (2a² - (C + A))sin2w.
• Thus a² = (B/sin2w + (C + A))/2.
• Since sin2w = B/�ˆš((C-A)² + B²),
• a² = [(C + A) + �ˆš((C - A)² + B²)]/2
• b² = C + A - a² = C + A - (�ˆš((C-A)² + B²) + (C + A))/2 and
• b² = [(C + A) - �ˆš((C - A)² + B²)]/2

What will these be in terms of the normalized equations? Substituting A=FA', etc., we have

• a² = [(FC' + FA') + �ˆš((FC' -FA')² + FB'²)]/2 or
• a² = F[(C' + A') + �ˆš((C' - A')² + 'B²)]/2 and similarly,
• b² = F[(C' + A') - �ˆš((C' - A')² + 'B²)]/2

All that is left is to find the center of the ellipse (h,k). We have

• D = - 2h[b²cos²w + a²sin²w] - 2k(a² - b²)sinwcosw
• E = - 2k[b²sin²w + a²cos²w] - 2h(a² - b²)sinwcosw

Since B = (a² - b²)sin2w = 2(a² - b²)sinwcosW, we can see that:

• D = - 2h[b²cos²w + a²sin²w] - kB
• E = - 2k[b²sin²w + a²cos²w] - hB

Also, A = [b²cos²w + a²sin²w] and C = [b²sin²w + a²cos²w]. Hence:

• D = - 2hA - kB
• E = - 2kC - hB

Not only do we have two simultaneous equations in h and k, they're gratifyingly neat equations as well. We can eliminate the k terms by multiplying the top equation by -2C and the bottom equation by B:

• -2CD = 4hCA + 2kCB
• BE = -2kCB - hB²
• We add the equations: -2CD + BE = 4hCA - hB²
• h = (BE - 2CD)/(4CA - B²)

Going back to the formula for D, we can solve for k

• k = -(D + 2hA)/b
• k = -(D + 2A(BE - 2CD)/(4CA - B²))/b
• k = -(D(4CA - B²) + 2A(BE - 2CD)/((4CA - B²)B)
• k = -(4CDA -DB² + 2ABE - 4ACD)/((4CA - B²)B)
• k = -(-DB² + 2ABE)/((4CA - B²)B)
• k = (DB - 2AE)/(4CA - B²)

Interestingly enough, since both h and k are second degree in both numerator and denominator, the scaling factor F plays no role.

Finally, the constant term
F = h²[b²cos²w + a²sin²w] + 2(a² - b²)(hk)sinwcosw + k²[b²sin²w + a²cos²w] - a²b²

We know

• A = [b²cos²w + a²sin²w]
• B = 2(a² - b²)sinwcosw
• C = [b²sin²w + a²cos²w]
• From above, we found that
  • a² = [(C + A) + �ˆš((C - A)² + B²)]/2
  • b² = [(C + A) - �ˆš((C - A)² + B²)]/2
  • Hence a²b²  = [(C + A)²  - ((C - A)² + B²]/4
  • a²b²  = [C² + 2AC +  A²  - C² + 2AC - A² - B²]/4
  • a²b²  = [4AC -  B²]/4 = AC - B²/4
• Thus, F = h²C + (hk)B + k²A - AC + B²/4
• F = h²FC' + (hk)FB' + k²FA' - FA'FC' + (FB')²/4
• 1 = h²C' + (hk)B' + k²A' - FA'C' + F(B')²/4
• F = (1 - h²C' + (hk)B' + k²A')/((B')²/4 - A'C')

Geometric Method

 

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Created 17 June 2005, Last Update 24 May 2020