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 [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?

 

[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.