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kelangan lang sa school... takot ko kay ma'am bataller eh

2006-07-10 00:44:40 · 7 answers · asked by Jenielle Gabriel 1 in Science & Mathematics Mathematics

7 answers

Parabola with vertical symmetry:
y = a(x - h)² + k
vertex: (h, k)
axis of symmetry: x = h
The parabola opens upward when a > 0, downward when a < 0.

Parabola with horizontal symmetry:
x = a(y - k)² + h
vertex: (h, k)
axis of symmetry: y = k
The parabola opens to the right when a > 0, to the left when a < 0.

2006-07-10 00:55:51 · answer #1 · answered by Anonymous · 0 0

y=x^2

2006-07-10 00:48:39 · answer #2 · answered by Shary M 2 · 0 0

Equations

(with vertex (h, k) and distance p between vertex and focus - note that if the vertex is below the focus, or equivalently above the directrix, p is positive, otherwise p is negative; similarly with horizontal axis of symmetry p is positive if vertex is to the left of the focus, or equivalently to the right of the directrix)
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Cartesian
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Vertical axis of symmetry

(x - h)^2 = 4p(y - k) \,

y = a(x-h)^2 + k \,

y = ax^2 + bx + c \,

\mbox{where }a = \frac{1}{4p}; \ \ b = \frac{-h}{2p}; \ \ c = \frac{h^2}{4p} + k; \ \
h = \frac{-b}{2a}; \ \ k = \frac{4ac - b^2}{4a}.

x(t) = 2pt + h; \ \ y(t) = pt^2 + k \,

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Horizontal axis of symmetry

(y - k)^2 = 4p(x - h) \,

x = a(y - k)^2 + h \,

x = ay^2 + by + c \,

\mbox{where }a = \frac{1}{4p}; \ \ b = \frac{-k}{2p}; \ \ c = \frac{k^2}{4p} + h; \ \
h = \frac{4ac - b^2}{4a}; \ \ k = \frac{-b}{2a}.

x(t) = pt^2 + h; \ \ y(t) = 2pt + k \,

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Semi-latus rectum and polar coordinates

In polar coordinates, a parabola with the focus at the origin and the top on the negative x-axis, is given by the equation

r (1 - \cos \theta) = l \,

where l is the semi-latus rectum: the distance from the focus to the parabola itself, measured along a line perpendicular to the axis. Note that this is twice the distance from the focus to the apex of the parabola or the perpendicular distance from the focus to the latus rectum.

2006-07-10 00:53:34 · answer #3 · answered by Anonymous · 0 0

The equation is y^2=4*a*x.

2006-07-10 00:49:02 · answer #4 · answered by rajesh_gshl 1 · 0 0

a quadratic formula given it is concave up or concave down this is how to get the equations.

y = a(x-h)^2 + k
where point (h,k) is the vertex h is the number of units from the y axis and k is the number of units from the x axis

2006-07-10 02:29:50 · answer #5 · answered by Croasis 3 · 0 0

You can get everything about Parabola in the following link

http://en.wikipedia.org/wiki/Parabola

2006-07-10 01:01:01 · answer #6 · answered by Sherlock Holmes 6 · 0 0

y=x sq n d vice versa

2006-07-10 02:41:58 · answer #7 · answered by anu 1 · 0 0

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