• Jul 25th 2008, 09:39 AM
norwoodjay
Write an quation for each parabola with vertex at orgin.

vertex (2,1), focus (-2,-3)
• Jul 25th 2008, 10:33 AM
earboth
Quote:

Originally Posted by norwoodjay
Write an quation for each parabola with vertex at orgin.

I don't understand this sentence :confused:

Quote:

vertex (2,1), focus (-2,-3)
If the parabola has it's vertex at (2,1) and it's focus at (-2,-3) then it has the equation:

p: x^2 - 2xy + y^2 + 30x + 34y = 95
• Jul 25th 2008, 11:00 AM
masters
Quote:

Originally Posted by earboth
I don't understand this sentence :confused:

If the parabola has it's vertex at (2,1) and it's focus at (-2,-3) then it has the equation:

p: x^2 - 2xy + y^2 + 30x + 34y = 95

How's that a parabola? Looks a lot like a circle to me.
• Jul 26th 2008, 01:44 AM
earboth
Quote:

Originally Posted by masters
How's that a parabola? Looks a lot like a circle to me.

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.
• Jul 26th 2008, 06:14 AM
masters
Quote:

Originally Posted by earboth
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.

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.