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

[chaoseverlasting]

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.

 

[arildno]

Do the shift of coordinates first, then worry about the eccentricity!

 

[chaoseverlasting]

How would you do that? By substitituting x+a for x and y+b for y to eliminate the xy term?

 

[arildno]

NO!
First of all, sorry for saying "shifting" the coordinates, I meant "rotating" the coordinates.

Do you know how to do that?

 

[chaoseverlasting]

Using x=xcos(t) - ysin(t) and y=xsin(t) + ycos(t) then equating the coeff of the xy term to zero from which you would get the value of tan2(t).

Then, substituting the value of sint and cost, you would get the general equation of the ellipse... right?

 

[arildno]

Right!
Then use, for example, the relation between the semi-major and semi-minor axes and the eccentricity to determine the latter quantity.

 

[chaoseverlasting]

Thank you, that helps a lot. Its great having such talented people there to look at your problems. Thanks a lot.