conic equations?

Question

What kind of a conic equation is this (parabola, circle, ellipse, hyperbola?):

-xy + 3y^2 - 4x + 2y + 8 = 0

Answer 1

If I assumed the equation is:

-x^2 + 3y^2 - 4x + 2y + 8 = 0

That would be a hyperbola.

Since it were -xy, it wouldn't have been any of them.

You need to seperate the x and y terms to either be a circle, parabola, ellipse, or hyperbola. At least I believe so.

Answer 2

If the equation is ax^2 + bxy + cy^2 + dx + ey + f = 0, compute b^2 - 4ac = D. The conic section is an ellipse if D < 0, a parabola if D = 0, and a hyperbola if D > 0. Here,
D = 1, so this is a hyperbola.

To eliminate the xy term, we would do a rotation of axes through an angle =
(1/2)*arctan (1/3). We could then put the new equation into standard form. The work is tedious, but the D-test easily lets us identify the curve.

Answer 3

That's not a conic section at all because for the
Circle is (x-a)^2 + (y-b)^2 =r^2
Parabola is Ax^2 +Bxy +Cy^2 +Dx +Ey+ F=0 where B^2 =4AC
Hyperbola A x^2 + B xy + C y^2 + D x + E y +F = 0
where B^2 >4AC and
Ellipse is [x/a]^2 + [y/b]^2 =1

Answer 4

4x²-5y²-16x-30y-9=0 4x²-16x-5y²-30y=9 4(x²-4x)-5(y²-6y)=9 4(x²-4x+4)-5(y²-6y+9)=9+4(4)+5(9) ---finished the sq. 4(x-2)²-5(y-3)²=86 (x-2)² (y-3)² ____ - ____ =a million 21.5 17.2 as a result this could be a hyperbola center(2,3)