conic sections: circles, ellipses, parabolas, hyperbolas

distance formula:

(x1-x2)^2 + (y1-y2)^2 = d^2 where d= the radius from point 1 to point 2

equation of a line:

ax + by = c

circle equation:

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

Given: c=(3,-2) and passes through (-1,1) solve it as a circle equation:

1. solve to find r: (-1-3)^2 + (1+2)^2 = 25 .... (x-3)^2 + (y+2)^2 = 25

If given the endpoints of diameter, use the midpoint equation:

((x1+x2)/2 , (y1+y2)/2) = center of the circle/ellipse

ellipse equation:

(((x-h)^2)/a^2) + (((y-k)^2)/b^2) =1

focal distance definition:

c: the distance from each focus to the center

focal distance equation for ellipses:

c^2 = a^2 - b^2

focal distance equation for hyperbolas:

c^2 = a^2 + b^2

eccentricity equation:

e = c ÷ (1/2 major axis)

in conic sections, vertices are...

the ends of the major axis

in parabolas when the y term is squared...

you know that it will be a sideways parabola.

y term squared equation for parabolas:

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

in sideways parabolas what happens if: 1. p<0 2.p>0

1. if p is less than 0 then the parabola will face left and be negative 2. if p is more than 0 then it will face right

parabola equations (2):

(x-h)^2 = 4p(y-k) <or> (y-k)^2 = 4p(x-h)

find the equation for the parabola. Given: vertex=(5,3) and points given on the parabola are (4.5,4) and (4.5,2).

since we don't know p, solve for it by inserting one of the points as (x,y)

1. hyperbola equation 2. hyperbola "c" equation

1. (((x-h)^2)/a^2) - (((y-k)^2)/b^2) =1 for a sideways parabola, or (((y-k)^2)/b^2) - (((x-h)^2)/a^2) for a vertical parabola 2. c^2 = a^2 + b^2 3.

Asymptotes equation:

y= + or - b/a (x-h) + k

hyperbola graphing: what way do you draw the curves if theres an -x^2 or a -y^2 in the equation?

-x^2 = veering away from the x axis, so vertical. -y^2= veering away from the y axis, horizontal

in changing a hyperbola equation from expanded form to standard form, you always...

factor out a negative coefficient which will change the sign in the parenthesis set.

1. vertices in hyperbolas are (definition)
2. Foci move between (in hyperbolas)

1. the ends of the hyperbola
2. hyperbolas, not along major axis of box

when balancing an equation...

you have to add/subtract the extra # on the outside of the parenthesis to both sides

hyperbolas foci equation:

F= (h, K + or - c)

hyperbolas vertices equation:

v= (h, k + or - b)

0 solution in hyperbolas:

degenerate cone, not solvable. when (h,k) is known, that's where the asymptotes intersect

How do you solve for b when given: foci= (0, + or - 8) Asymptotes: y= + or - 4x

c= 8 so 8 = √a^2 + b^2
b/a = 4 so b= 4a. Then insert 4a for b so the equation becomes 8 = √a^2 + 4a^2 and use algebra to solve for a. Then insert it back in to solve for b.

.

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if only one term is squared, then it must be a...

parabola

if both x and y are squared but only one is positive, what is it?

hyperbola

if both x and y are squared and both are positive and it has the same coefficient then it is a...

circle

if both x and y are squared and both are positive and it has different coefficients then it is a...

ellipses

ellipses eccentricity number has to be between...

0<e<1

what things do you have to find for a circle

center, radius

what things do you have to find for an ellipse

center, foci, vertices, eccentricity

what things do you have to find for a hyperbola

center, foci, vertices, asymptotes

what things do you have to find for a parabola

vertex, focus, directrix, intercepts

What is p?

the distance between the vertex and the focus, and the distance between the vertex and the directrix

What do you do if it asks for an extra point for graphing?

you use an intercept. For example, if it was a sideways equation, you use the point where y=0 and solve for x to get the point.

Turn this parabola equation from expanded to standard form: x^2 + 2x -9y + 28 = 0
Then, what's the vertex?

1(x^2 + 2x + 1) = -28 +9y + 1
(x + 1)^2 = 9(y - 3)
9= 4p so p=9/4 = 2 1/4
vertices (1, -3) not (1, -27)