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**Soroban** Hello, ladventchildp!

If you're doing Rotations, you must be familiar with certain formulas.

The general quadratic equation is: .Ax² + Bxy + Cy² + Dx + Ey + F .= .0

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B

The angle of rotation θ is given by: .tan(2θ) .= .------

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .A - C

. . . . . . . . . . . . . . . . . . . _

We have: .A = 31, B = 10√3, C = 21, D = 0, E = 0, F = -144

. . - . . - . . . . . . . . . . . _

. . . . . . . . . . . . . . .10√3 . - - . _

Hence: . tan2θ .= .--------- .= .√3

. . . . . . . . . . . . . .31 - 21

Then: .2θ .= .π/3 . . → . . **θ .= .π/6**

To eliminate the xy-term, we use these substitutions:

. . x' .= .x·cosθ - y·sinθ

. . y' .= .x·sinθ + y·cosθ

With θ = π/6, we have:

. . x .= .½√3·x - ½·y

. . y .= .½·x + ½√3·y

Substitute:

31(½√3x - ½y)² + 10√3(½√3x - ½y)(½x + ½√3y) + 21(½x + ½√3y)² - 144 .= .0

This simplifies to: .36x² + 16y² .= .144

. . . . . . . . . . . . . . . . . .x² . . y²

The conic is an ellipse: . --- + --- .= .1

. . . . . . . . . . . . . . . . . .4 . . .9

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We could have identitied the conic long ago.

The equation has a **discriminant**: .d .= .B² - 4AC . . (Look familiar?)

. . If d = 0: parabola

. . If d > 0: hyperbola

. . If d < 0: ellipse

. . . . . . . . . . . . . . . . . . . _

We have: .A = 31, B = 10√3, C = 21

. . - . . . - . . . - . . . . _

Therefore: .d .= .(10√3)² - 4(31)(21) .= .-2304 . . . . ellipse