SOLUTION: Use the rotation formulas to rotate the ellipse to standardized form. Ellipse: {{{ x^2+xy+y^2=6 }}}

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Question 1175625: Use the rotation formulas to rotate the ellipse to standardized form.
Ellipse:
+x%5E2%2Bxy%2By%5E2=6+

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
The conic section

Ax%5E2+%2B+Bxy+%2B+Cy%5E2+%2B+Dx+%2B+Ey+%2B+F+=+0  

when rotated through angle θ becomes this equation:

Gu%5E2+%2B+Huv+%2B+Kv%5E2+%2B+Lx+%2B+My+%2B+N+=+0

where cot%282theta%29=%28A-C%29%2FB

Ordinarily instead of using different letters most math people use primes to
the right of the same letters, but the primes confuse me because they go up
where exponents normally go. So to keep from getting confused, I use different
letters and no primes.

G+=+A%2Acos%5E2%28theta%29+%2B+B%2Asin%28theta%29cos%28theta%29+%2B+C%2Asin%5E2%28theta%29 
H+=2%28C-A%29sin%28theta%29cos%28theta%29+%2B+B%28cos%5E2%28theta%29+-+sin%5E2%28theta%29%29
K+=A%2Asin+%5E2%28theta%29+-+B%2Asin%28theta%29cos%28theta%29+%2BC%2Acos%5E2%28theta%29
L+=+D%2Acos%28theta%29+%2B+E%2Asin%28theta%29 
M+=+-Dsin%28theta%29+%2B+Ecos%28theta%29 
N+=+F 

In your problem,

+x%5E2%2Bxy%2By%5E2=6+ can be written in form

Ax%5E2+%2B+Bxy+%2B+Cy%5E2+%2B+Dx+%2B+Ey+%2B+F+=+0 as

1x%5E2%2B1xy%2B1y%5E2%2B0x%2B0y-6=0+

A=1, B=1, C=1, D=0, E=0, F=-6

cot%282theta%29=%28A-C%29%2FB=%281-1%29%2F1=0%2F1=0
So
2theta+=+matrix%281%2C3%2C90%5Eo%2Cor%2C270%5Eo%29
theta+=+matrix%281%2C3%2C45%5Eo%2Cor%2C135%5Eo%29

We could choose to rotate it through either of those, but 45o is 
the simplest.

 

K+=1%2Asin%5E2%2845%5Eo%29+-+1%2Asin%2845%5Eo%29cos%2845%5Eo%29+%2B1%2Acos%5E2%2845%5Eo%29
L+=+0%2Acos%2845%5Eo%29+%2B+0%2Asin%2845%5Eo%29 
M+=+-0sin%2845%5Eo%29+%2B+0cos%2845%5Eo%29 
N+=+-6

Simplifying:

G=3%2F2
H=0
K=1%2F2
L=0
M=0
N=-6

So

Gu%5E2+%2B+Huv+%2B+Kv%5E2+%2B+Lx+%2B+My+%2B+N+=+0

becomes:

expr%283%2F2%29u%5E2+%2B+0uv+%2B+expr%281%2F2%29v%5E2+%2B+0x+%2B+0y+-+6+=+0

expr%283%2F2%29u%5E2+%2B+expr%281%2F2%29v%5E2+-+6+=+0

3u%5E2+%2B+v%5E2+-+12+=+0

3u%5E2+%2B+v%5E2+=+12

Divide through by 12

3u%5E2%2F12%5E%22%22+%2B+v%5E2%2F12%5E%22%22+=+12%5E%22%22%2F12%5E%22%22

u%5E2%2F4%5E%22%22%2Bv%5E2%2F12%5E%22%22=1

That's an ellipse of the form:

u%5E2%2Fb%5E2%2Bv%5E2%2Fa%5E2=1

Since 12 is larger than 4, the ellipse is vertical
with respect to the u-v axis, with center (0,0),
semi-major axis √12 = 2√3 and semi-minor axis √4 = 2



If we had used 135o instead, the u and v axes would have been
switched but the graph would have been the same.

Edwin