## Wednesday, May 27, 2009

### Is it a parabola? is it an ellipse? no its.. BOB

Well this unit was pretty short and sweet and fairly simple. I'm going to get right into a review because I've got to work on DEV.

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A parabola is a locus of points a fixed distance from a fixed point (focus) and a line (directrix).

Vertical Parabola: (x-h)^2=4p(y-k)
Horizontal Parabola: (y-h)^2=4p(x-k)

The coordinates (h,k) are the vertex of the parabola. "p" is the distance from the vertex to the focus or the vertex to the directrix.  These distances are both equal. When 4p is greater than one, the parabola is wider and when it is less than one, the parabola is narrower.

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A circle is the locus of points equidistant from a fixed point, the center and the distance to any one point is called the radius.

Circle: (x-h)^2 + (y-k)^2= r^2

The coordinates (h,k) are the vertex of the circle. "r" is the radius of the circle.

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An ellipse is the locus of points which the sum of the distances to the foci is known.

Horizontal Ellipse: [(x-h)^2/a^2] + [(y-k)^2/b^2] = 1
Vertical Ellipse: [(y-h)^2/a^2] + [(x-k)^2/b^2] = 1

In an ellipse addition is used because the locus of points is the sum of the distances. The center of the ellipse is the coordinates (h,k) and is halfway between the two foci. The semi major axis is equal to "a" and the whole major axis is equal to "2a". The semi minor axis is equal to "b" and the whole minor axis is equal to "2b". The distance of either foci to the center is distance "c" and can be determined by c^2=a^2-b^2.

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A hyperbola is the locus of points which the difference of the distances to the foci is known

Horizontal Hyperbola: [(x-h)^2/a^2] - [(y-k)^2/b^2] = 1
Vertical Hyperbola: [(y-h)^2/a^2] - [(x-k)^2/b^2] = 1

The transverse axis is the distance between the vertices of the hyperbola. This is distance "2a". The distance from either vertex to the center is "a" and is the semi transverse axis. The conjugate axis is perpendicular to the transverse axis and it's distance is "2b". Half the distance, "b", is called the semi conjugate axis. The distance of the foci can be calculated c^2= a^2+b^2.

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Well that is all so I'm to work on DEV! I hope everyone did well on the test!