Question:
How Does One Locate a Tangent on an Ellipse?
Dico
2012-07-29 16:45:48 UTC
I'm drawing an object in AutoCAD 2010 involving ellipses. In particular, I have a troublesome ellipse rotated -15 degrees, and I need to find the lowest point -- in other words, I need to find a point where the tangent line is parallel to the x-axis.

If it helps (I don't need coordinates if you can provide detailed instructions, but if you don't mind, coordinates are always better):
Centre - (148.281 , 19.2309)
Major Axis - 19.16
Minor Axis - 11.48
Direction - Major Axis is 'pointing' to 165 degrees and 345 degrees

I'm open to both math-related solutions and solutions specific to AutoCAD (however, I will not use the 'pellipse' command). That said, simpler is better: I'm only sixteen, so Calculus isn't preferable. I tried it with another tangent-ellipse issue and had quite a bit of trouble.

In advance, thanks.
Four answers:
Jenna K
2012-07-30 19:59:49 UTC
It does take calculus to solve this, but let's see if we can do the work up front to turn this into a formula you can use for any ellipse.



Start with standard form: x^2 / a^2 + y^2 / b^2 = 1, where 2a and 2b are the axes. Let's use the x axis as the major axis. Now, to rotate this θ degrees, we use the transformation

(x,y) -> (x cos θ + y sin θ, -x sin θ + y cos θ)



Plug this in to the ellipse equation:



(x cos θ + y sin θ)^2 / a^2 + (-x sin θ + y cos θ)^2 / b^2 = 1

b^2(x cos θ + y sin θ)^2 + a^2 (-x sin θ + y cos θ)^2 = a^2 b^2



b^2(x^2 cos^2 θ + 2xy cos θ sinθ + y^2 sin^2 θ) + a^2 (x^2 sin^2 θ - 2xy cos θ sinθ + y^2 cos^2 θ) = a^2 b^2



(b^2 cos^2 θ + a^2 sin^2 θ) x^2 + (b^2 - a^2)(2 cos θ sin θ) xy + (b^2 sin^2 θ + a^2 cos^2 θ) y^2 = a^2 b^2



(b^2 cos^2 θ + a^2 sin^2 θ) x^2 + (b^2 - a^2)(2 cos θ sin θ) xy + (b^2 sin^2 θ + a^2 cos^2 θ) y^2 = a^2 b^2





This looks complicated, but for any ellipse where you know the axes and the degree of rotation, this becomes a reasonably simple equation. Let's not plug numbers in just yet, though. First, differentiate, treating a, b, and θ as constants:



2(b^2 cos^2 θ + a^2 sin^2 θ) x + (b^2 - a^2)(2 cos θ sin θ) (xy' + y) + 2(b^2 sin^2 θ + a^2 cos^2 θ) y y' = 0



Set y' to 0:

2(b^2 cos^2 θ + a^2 sin^2 θ) x + (b^2 - a^2)(2 cos θ sin θ)y= 0

y = [-2(b^2 cos^2 θ + a^2 sin^2 θ) / (b^2 - a^2)(2 cos θ sin θ)] x



Since a, b, and θ are constant for a given ellipse, this is a line. It will intersect with the ellipse at the highest and lowest points, where the derivative is zero.



So, for any ellipse, you can use the following steps to find the maximum and minimum without calculus, since we've already done that part of the work.



a = half the major axis = 9.58

b = half the minor axis = 5.74

θ = degree of rotation = 165



Plug into the ellipse equation:

(b^2 cos^2 θ + a^2 sin^2 θ) x^2 + (b^2 - a^2)(2 cos θ sin θ) xy + (b^2 sin^2 θ + a^2 cos^2 θ) y^2 = a^2 b^2



36.8884 x^2 + 29.4144 xy + 87.8356 y^2 = 3023.81



Plug into the line equation:

y = [-2(b^2 cos^2 θ + a^2 sin^2 θ) / (b^2 - a^2)(2 cos θ sin θ)] x

y = -2.50819 x



Substitute y from the line equation into the ellipse equation:

36.8884 x^2 + 29.4144 x*(-2.50819 x) + 87.8356 (-2.50819 x)^2 = 3023.81

515.685 x^2 = 3023.81

x = +/- 2.42150

y = -2.50819 * (+/- 2.42150) = -/+ 6.07358



We have found two points: (2.42140, -6.07358) and (-2.42140, 6.07358). These are the highest and lowest points on an ellipse that is the same size and orientation as yours, but centered on the origin. All you need to do now is add your center coordinates to shift these to their proper location.



(2.42140 + 148.281, -6.07358 + 19.2309) = (145.860, 25.3045)

(-2.42140 + 148.281, 6.07358 + 19.2309) = (150.703, 13.1573)



Edit: I did a little more algebra; here's a different method you may find easier to use.



The ellipse equation (b^2 cos^2 θ + a^2 sin^2 θ) x^2 + (b^2 - a^2)(2 cos θ sin θ) xy + (b^2 sin^2 θ + a^2 cos^2 θ) y^2 = a^2 b^2



can be rewritten as

A x^2 + B xy + C y^2 = F

where

A = b^2 cos^2 θ + a^2 sin^2 θ

B = (b^2 - a^2)(2 cos θ sin θ)

C = b^2 sin^2 θ + a^2 cos^2 θ

F =a^2 b^2



Now looking at the line equation, it turns out that this is

y = (-2A / B) x



Substituting for y gives

A x^2 + B (-2A / B) x^2 + C(-2A/B)^2x^2 = F



which can be rearranged to

x = +- B sqrt(F / (4A^2 C - AB^2) )



And therefore

y = -+ 2A*(sqrt(F / (4A^2 C - AB^2) )



So, for any given ellipse, the steps become:



1. Find A, B, C, and F using the equations above.

2. Find K = sqrt(F / (4A^2 C - AB^2)

3. x = +- BK, and y = -+ 2AK

4. Add your center coordinates to the x and y calculated in step 3.
Bong
2012-08-01 04:46:58 UTC
You can obtain using a smple solution in Autocad, as accurate as 4 decimal units.



You should be able to point out the lowest point on the ellipse roughly.



1. Draw a vertical line crossing that part of the curve where the lowest point is somewhere located.

2. Offset the line 0.01 units left or right until you think the target point is covered. Use "Multiple" option in offset.

3. Use the ellipse to trim all those lines leaving the upper parts.

4. Check the measurement of each line. The longest should be crossing the lowest point.



Note: Configure Autocad to display the Length property of the line when you hover it with the cursor. Or you can enable the "quick properties" on the task bar.
CogitoErgoCogitoSum
2012-07-29 17:04:06 UTC
Calculus is the only mathematical way of solving this. It's either that or you use the built in commands at your disposal.



I guarantee that any command on autocad makes use of calculus behind the scenes
giniebra
2016-12-10 15:04:24 UTC
(x-5)^2 / a^2 + (y+a million)^2/ b^2 =a million a is the path left and properly suited from the midsection to the perimeters. in spite of if that is tangent to the y-axis , that ought to represent a=5 b is the path up and down from the midsection to the perimeters. in spite of if that is tangent to the x-axis, then b=a million. So, (x-5)^2 / 25 + (y+a million)^2/ a million =a million


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