How Do You Calculate the Square of Eccentricity for a Rotated Ellipse?

In summary, the conversation discusses the problem of finding the square of the eccentricity of an ellipse represented by the equation ax^2 + 2hxy + by^2 =1. The participants suggest using the ratio of the distance from the directrix to the focus of a point on the ellipse to find the eccentricity, as well as rotating the coordinates and using the relation between the semi-major and semi-minor axes to determine the eccentricity. They also express gratitude for the help and expertise of others in solving the problem.
  • #1
chaoseverlasting
1,050
3
This one question has me totally beaten. And I thought I was pretty good in co-ordinate geometry. Here it is:

If the equation ax^2 + 2hxy + by^2 =1 represents an ellipse, find the square of the eccentricity of the ellipse.

I know that the ratio of the distance from the directrix to the focus of a point on the ellipse is the eccentricity. But I can't figure out what the directrix is or where the foci lie. This equation must represent an ellipse with its axes shifted (as the equation with x and y axes as its major axes is (x^2/a*a) + (y*y/b*b) =1). Also, here h*h - ab <0, and abc +2fgh -af*f - bg*g -ch*h is non zero. I just don't know how to go about finding the eccentricity.
 
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  • #2
Do the shift of coordinates first, then worry about the eccentricity!
 
  • #3
How would you do that? By substitituting x+a for x and y+b for y to eliminate the xy term?
 
  • #4
NO!
First of all, sorry for saying "shifting" the coordinates, I meant "rotating" the coordinates.

Do you know how to do that?
 
  • #5
Using [tex] x=xcos(t) - ysin(t)[/tex] and [tex] y=xsin(t) + ycos(t) [/tex] then equating the coeff of the xy term to zero from which you would get the value of [tex]tan2(t)[/tex].

Then, substituting the value of sint and cost, you would get the general equation of the ellipse... right?
 
  • #6
Right!
Then use, for example, the relation between the semi-major and semi-minor axes and the eccentricity to determine the latter quantity.
 
  • #7
Thank you, that helps a lot. Its great having such talented people there to look at your problems. Thanks a lot.
 

Related to How Do You Calculate the Square of Eccentricity for a Rotated Ellipse?

1. What is the eccentricity of an ellipse?

The eccentricity of an ellipse is a measure of how "elongated" or "squashed" the ellipse is. It is represented by the letter e and is defined as the ratio of the distance between the foci of the ellipse to the length of its major axis.

2. How is the eccentricity of an ellipse calculated?

The eccentricity e of an ellipse can be calculated using the formula e = c/a, where c is the distance between the foci and a is the length of the major axis. Alternatively, it can also be calculated using the formula e = √(1 - b2/a2), where b is the length of the minor axis.

3. What is the range of values for the eccentricity of an ellipse?

The eccentricity of an ellipse can range from 0 to 1. A value of 0 represents a circle, while a value of 1 represents a parabola. Any value between 0 and 1 represents an ellipse of varying degrees of "squashing".

4. What is the significance of the eccentricity of an ellipse?

The eccentricity of an ellipse is an important parameter in understanding the shape and properties of an ellipse. It affects the size, orientation and curvature of the ellipse, as well as the locations of its foci and vertices. It is also used in applications such as celestial mechanics and orbital mechanics to describe the orbits of objects.

5. Can the eccentricity of an ellipse be greater than 1?

No, the eccentricity of an ellipse cannot be greater than 1. This would result in an imaginary value for the length of the major axis, which is not possible in a real world scenario. A value of 1 or greater would also result in a non-closed curve, such as a hyperbola, instead of an ellipse.

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