Write an quation for each parabola with vertex at orgin.

vertex (2,1), focus (-2,-3)

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- July 25th 2008, 10:39 AM #1

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- July 25th 2008, 11:33 AM #2

- July 25th 2008, 12:00 PM #3

- July 26th 2008, 02:44 AM #4
I've attached the diagram of the parabola.

From the given points you can calculate the equation of the axis, the directrix d and the coordinates of the points B and C.

EDIT: If you have the general equation of a conic:

Ax² + 2Bxy + Cy² + 2Dx + 2Ey + F = 0

then you can determine which conic you have by calculating

AC - B² > 0 ----> ellipse

AC - B² = 0 ----> parabola

AC - B² < 0 ----> hyperbola

You'll see that the coefficients of the equation I posted satisfy the condition indicating a parabola.

- July 26th 2008, 07:14 AM #5

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Thanks, Earboth

I haven't worked that much with transformations of conics. I knew the axis of symmetry was oblique, but had forgotten about identifying the conic by using the discriminant.

So this parabola is a rotation about the origin, and the angle of rotation is pi/2 (45 degrees) since A = C.

I am more familiar with: Ax^2 + Bxy + Cy^2 + Dx +Ey + F = 0

as the general equation. Then, testing the discriminant (B^2 - 4AC),

If B^2 - 4AC < 0, and A = C, then it's a circle. If A not equal C, then it's an ellipse

If B^2 -4AC > 0, then it's a hyperbola

If B^2 - 4AC = 0, then it's a parabola.

I have to think the OP made a typo on this one, but it turned out to be a great post....for me, anyway.