unit 3

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circle =

0 eccentricity

for eccentricity values

0 is less than or equal to eccentricity value which is less than 1

how to graph hyperbolas: once the equation is in standard form and you have determined the values of a, b and c. follow the steps below to graph:

1. use "a" to plot the vertices on the major (or transverse) axis 2. use "b" to create a rectangle around the ends of all axes 3. sketch the diagonals of this rectangle. these are the asymptotes. write their equations in slope-intercept or point-slope form depending on the center 4. use the asymptotes to graph the curve (pass through the vertices) 5. use "c" to plot the foci on the major axis. they should fall within the curve

minor axis value

smaller denomenator

write the equation of the circle or ellipse in conics standard form: 4x² + y² + 8x - 6y - 3 = 0 identify which type of conic (circle or ellipse) you have: state the coordinates of the center of the conic

((x + 1)² / 4) + ((y - 3)² / 16) = 1 ellipse center: (-1,3)

write in conics form. state the center. find its eccentricity. 2x² + 9y² + 4x - 18y - 7 = 0

((x + 1)² / 9) + ((y - 1)² / 2) = 1 center: (-1,1) eccentricity: square root of 7 / 3

find standard form, graph, and find asymptote 25x² - 4y² + 100x + 24y - 36 = 0

((x + 2)² / 4) - ((y - 3)² / 25) = 1 center (-2,3) vertices (-4,3) (0,3) (y - 3) = +/-(5/2)(x + 2)

answer the following questions based on the graph: center: (1,-2) vertices: (4,-2) (-2,-2) (1,-1) (1,-3) foci: almost at the vertices of the horizontal line write an equation for the ellipse: find the eccentricity of the ellipse (round to 1 decimal place):

((x - 1)² / 9) + (y + 2)² = 1 0.9 very elliptical

write the equation of an ellipse with its center at (2,-4), minor axis with a total length of 20 and one of its foci at (2,16).

((x - 2)² / 100) + ((y + 4)² / 500) = 1

write the equation of the ellipse given the following formation: center: (3,5), and is tangent to both axes

((x - 3)² / 9) + ((x - 5)² / 25) = 1 a: 5 b: 3

find in standard form, graph, and find asymptote 5x² - 4y² - 40x - 16y - 36 = 0

((x - 4)² / 20) - ((y + 2)² / 25) = 1 center (4,-2) vertices (-.5,-2) (8.5,-2) (y + 2) = +/-(5/4.5)(x - 4)

standard equation for hyperbola: horizontal transverse axis

((x - h)² / a²) - ((y - k)² / b²) = 1 asymptotes cross/are on the horizontal axis

make equation: center at (1,1) vertex at (1,5) asymptotes with slope of +/-1/2 hyperbola

((y - 1)² / 16) - ((x - 1)² / 64) = 1

find the standard form, transverse axis, center, a, b, c², c, asymptotes, and graph it 9y² - 4x² - 36y + 8x - 4 = 0

((y - 2)² / 4) - ((x - 1)² / 9) = 1 vertical (1,2) 2 3 13 square root of 13 = 3.6 (y - 2) = +/-(2/3)(x - 1) vertices (1,0) (1,4)

write the equation of the hyperbola given that V1(4,0), V2(4,8), and asymptotes have a slope of +/-1

((y - 4)² / 16) - ((x - 4)² /16) = 1

standard equation for hyperbola: vertical transverse axis

((y - k)² / a²) - ((x - h)² / b²) = 1 asymptotes cross/are on the vertical axis

Equation of a circle in standard form/conic standard form/graphing form

(x - h)² + (y - k)² = r²

vertical axis of symmetry - equation: p is greater than 0: p is less than 0:

(x - h)² = 4p(y - k) opens up opens down

vertex

(h,k); half way between focus and directrix

write an equation for a circle with center at (2,-1) and radius of 9

(x - 2)² + (y + 1)² = 81

what is the exact distance between any point on your circle and the center of the circle? use this distance with your expression from part a to write an equation

(x - 3)² + (y - 5)² = 16 16 is the radius squared (4²)

standard conics form: vertex: p: focus: directrix: opens: y = -1/2x² + 4x - 9

(x - 4)² = -2(y + 1) (4,-1) -1/2 (4,-1 1/2) y = -1/2 down

write equation of parabola with vertex at (6,3) and focus at (6,8)

(x - 6)² = 20(y - 3)

a bridge is built in the shape of a semi-elliptical arch. the bridge has a span of 300 feet and a maximum height of 60 feet. choose a suitable coordinate system to write an equation for the ellipse in standard conics form: find the height of the arch at a distance of 15 feet from its center

(x² / 22500) + (y² / 3600) = 1 59.70ft

write the equation of the ellipse given the following formation: center: (0,0), foci: (3,0), and major axis of length 10

(x² / 25) + (y² / 16) = 1 a: 5 b: 4 c: 3 major axis is horizontal

write the equation of the ellipse given the following formation: center: (0,0), vertex: (0,-3), and semi-minor axis of length 2

(x² / 4) + (y² / 9) = 1 a: 3 b: 2 whole minor axis = 4

put in standard form, graph, find c value, find asymptote equations: 9x² - 4y² = 36

(x² / 4) - (y² / 9) = 1 middle c = square root of 13 y = +/- 3/2x

write the equation of the hyperbola given that C(0,0), V(8,0), and asymptote is y = 1/2x

(x² / 64) - (y² /16) = 1

find the equation of the ellipse with vertices at (9,0) and (0,-2) and center at (0,0)

(x² / 81) + (y² / 4) = 1

write in standard conics form: 2y - y² = x - 3

(y - 1)² = -1(x - 4)

write equation vertex (2,-1) directrix x = 1

(y - 1)² = -12(x + 2)

write equation of parabola with directrix x = -4 and focus at (8,10)

(y - 10)² = 24(x - 2)

put in standard form, graph, state c value, and asymptote equations: -x² + 4y² - 2x - 16y + 11 = 0

(y - 2)² - ((x + 1)² / 4) = 1 Q IV c = square root of 5 (y - 2) = +/- 1/2(x + 1)

write equation focus (2,3) directrix x = -2

(y - 3)² = 8x

write in standard conics form: x - 1 = (-1/20)(y - 4)²

(y - 4)² = -20(x - 1)

horizontal axis of symmetry - equation: p is greater than 0: p is less than 0:

(y - k)² = 4p(x - h) opens right opens left

write the equation of the hyperbola given that there is a center at (0,0), vertices at (0,-12), and foci at (0,-13)

(y² / 144) - (x² / 25) = 1

write the equation of the hyperbola given that there is a V(2,3) and (2,-3), and passes through (0,5)

(y²/ 9) - ((x - 2)² / 2.25) = 1

p: opens: vertex: focus: directrix: graph: equation: (y - 2)² = -4(x + 3)

-1 left (-3,2) (-4,2) x = -2 Q II

p: opens: vertex: focus: directrix: graph: equation: (x - 5)² = -20(y + 2)

-5 down (5,-2) (5,-7) y = 3 Q IV

for an ellipse that is nearly circular, the foci are close to the center and the ratio c/a is

0

p: opens: vertex: focus: directrix: graph: equation: (y - 4)² = 8(x - 3)

2 right (3,4) (5,4) x = 1 Q 1

foci for hyperbola

2 fixed points lie on transverse axis "c" away from the center

foci

2 fixed points on the major axis

hyperbolas have...

2 vertices, not 4

p: opens: vertex: focus: directrix: graph: equation: (x - 3)² = 12(y + 1)

3 up (3,-1) (3,2) y = -4 mainly in Q IV

directrix is v or h: axis of symmetry is v or h: coordinate of vertex: "p" value: equation of directrix: coordinate of focus: equation: x² = -16(y + 3)

H V (0,-3) -4 y = 1 (0,-7) southern hemisphere

directrix is v or h: axis of symmetry is v or h: coordinate of vertex: "p" value: equation of directrix: coordinate of focus: equation: (x - 3)² = 8y

H V (3,0) 2 y = -2 (3,2) mostly in Q 1

semi-minor axis

Half of the minor axis

Vertices for Hyperbola

ONLY 2! lie "a" away from center on transverse axis

directrix is v or h: axis of symmetry is v or h: coordinate of vertex: "p" value: equation of directrix: coordinate of focus: equation: (y + 2)² = -4x

V H (0,-2) -1 x = 1 (-1,-2) graph is in western hemisphere

directrix is v or h: axis of symmetry is v or h: coordinate of vertex: "p" value: equation of directrix: coordinate of focus: equation: (y + 4)² = 2(x - 4)

V H (4,-4) 1/2 x = 3.5 (4.5,-4) graph is in Q IV

major axis value

bigger denomenator

parabola defintion

a set of points that are equidistant from a fixed line (directrix) and a fixed point (focus) not on the line

elongated ellipse =

approaches 1 (can't have eccentricity of 1 or more than 1)

relationship between a, b, and c

a² = b² + c²

relationship between a, b, and c for ellipse

a² = b² + c² b is vertical leg, c is horizontal leg, and a is hypotenuse

find eccentricity and classify it: (x² / 4) + y² = 1

c = 1.73 a = 2 c/a = 0.87 very elliptical, more elongated in comparison to the previous example

find eccentricity and classify it: (x² / 8) + (y² / 3) = 1

c = 2.24 a = 2.83 c/a = 0.79 very elliptical, less elongated in comparison to the next example

in hyperbolas...

c is now the biggest distance c is greater than a a is to vertice c is to foci b is to other point on conjugate axis b value frames wideness/narrowness of the hyperbola

graph the hyperbola and find the asymptote ((y - 4)² / 25) - ((x + 4)² / 9) = 1

center (-4,4) vertices (-4,9) (-4,-1) (y - 4) = +/-(5/3)(x + 4)

graph the hyperbola and find the asymptote (x + 5)² - ((y + 2)² / 49) = 1

center (-5,-2) vertices (-6,-2) (-4,-2) (y + 2) = +/-7(x + 5)

graph each hyperbola and find the asymptote ((x + 7)² / 25) + (y + 8)² = 1

center (-7,-8) vertices (-12,-8) (-2,-8) (y + 8) = +/-(1/5)(x + 7)

graph each hyperbola and find the asymptote (y² / 25) - (x² / 36) = 1

center (0,0) vertices (0,5) (0,-5) y = +/-(5/6)x

graph the following conic section. include center and vertices on your graph. ((x - 2)² / 4) + (y + 1)² = 1

center (2,-1) vertices (0,-1) (2,0) in QIV

in a circle, the foci =

center (on top of each other)

now draw a circle with center (3,5) and radius 4 units. again label a point P with the coordinates (x,y)

center of (3,5) radius of 4 all around this } rotated to the right (horizontal opening up) x = 3 this } y = 5 P (x,y)

c =

center to focus

a =

center to vertex of major axis

when the foci are extremely close to the..., it is difficult to make a visual distinction between a ..... and ....

center; circle; an ellipse

types of conics

circle, ellipse, hyperbola, parabola

closer to 0 is more

circular

look at applications in parabola packet

cover do check

look at problem 3.2 hyperbola applications

cover do check

look at 3.1 ellipse application

cover and do and check

look through all

cover and do and check

relationship between a, b, or c in a hyperbola

c² = a² + b²

a² is always under the...

major axis

p value

distance from V to F and V to D (separate)

asymptotes

drawn through the center and corners of the "box" only use y = mx + b when the center = (0,0) use point slope form: (y - y(subscript)1) = +/-m(x - x(subscript)1)

early astronomers had a difficult time when studying planetary orbits whose foci were relatively close to their center. now we can use the concept of

eccentricity

closer to 1 is more

elliptical

for an elongated ellipse, the foci are close to the ..... and the ratio c/a is close to

end point of the major axis; 1

vertices

endpoints of the axes

directrix

fixed line; "p" away from vertex

focus

fixed point; "p" away from vertex

4p value

focal width (width of parabola across the focus (in line with it (when given an application problem, you don't always know what the focus is so you can't just assume to focal width/4p value, you have to plug in other numbers))

the eccentricity (e) of an ellipse is...

given by the ratio: e = (c/a)

draw a circle with its center at the origin and a radius of 5. Mark a point P on the circle and label it with he coordinates (x,y)

graph radius of 5 all around p (x,y)

find transverse axis, center, a, c, c² = a² + b², c, asymptotes, and graph it ((x + 2)² / 9) - ((y - 1)² / 49) = 1

horizontal (-2,1) 3 7 58 square root of 58 = 7.62 (y-1) = +/-(7/3)(x + 2) vertices (-5,1) (1,1)

find transverse axis, a, b, c, c² = a² + b², c, asymptotes, and graph it (x² / 16) - (y² / 25) = 1

horizontal (because x came first (first numerator)) horizontal 4 5 16 + 25 = 41 c = square root of 41 = 6.4 asymptotes: y = +/-(5/4)x center: (0,0) vertices: (3,0) (-3,0) foci: (-6.4,0) (6.4,0)

2 cones = 1 cone =

hyperbola parabola

graph the following conic section. (x - 2)² + y² = 16

in right hemisphere center at (2,0)

center

intersection of 2 axes (h, k)

general form is when

it is all sprawled out and is not simplified

transverse axis

length of 2a "a" comes first in equation (first denominator) (is actually a²) vertices lie on transverse axis

conjugate axis

length of 2b "b" comes 2nd in the equation (2nd denominator) (b² in the equation) help frame the asymptotes

major axis

longer axis length = 2a

tangent to an axis

means that it touches it at one point but bounces back

b² is always under the...

minor axis

graph and find asymptote ((x - 2)² / 16) - ((y + 3)² / 4) = 1

more in Q IV (y + 3) = +/-1/2(x - 2)

standard form, graph, asymptotes: 4x² - 9y² - 40x + 64 = 0

more in eastern hemisphere ((x - 5)² / 9) - (y² / 4) = 1 y = +/-2/3(x - 5)

graph the following conic section ((x - 1)² / 25) + ((y + 2)² / 9) = 1

mostly in QIV center at (1,-2)

what happens as the focus moves nearer the directrix?

narrower parabola

vertical lines would be

on the sides

hyperbolas always...

open around its focus

ellipticalness/how elliptical of an ellipse aka

ovalness/how oval

when slicing a cone to find a circle you should...

parallel to the base of the cone

axis of symmetry

perpendicular to D through the V

notice that both forms of the standard equation of the ellipse are...

set equal to 1

ellipse definition

set of points the sum of 2 distances from foci is constant

hyperbola defintion

set of points, such that the difference from 2 fixed points (foci) is constant

minor axis

shorter axis length = 2b

when slicing a cone, to create an ellipse you should...

slice the cone diagonally, not through its base

when slicing a cone to find a parabola you should...

slice the cone parallel to the side of the cone

when slicing a cone to find a hyperbola you should...

slice the cone perpendicular to its base

put the following equations into standard conic form and sketch graph, find center, find a, find b, find the radius in both the x and y directions 9x² + 4y² + 90x - 16y + 205 = 0

standard conic form: ((x + 5)² / 4) + ((y - 2)² / 9) = 1 center: (-5,-2) a: 2 b: 3 radius in y direction: square root of 9 or 3 radius in x direction: square root of 4 or 2

put the following equations into standard conic form and sketch graph, find center, find a, find b 16x² + 25y² - 300y + 500 = 0

standard conic form: (x² / 25) + ((y - 6)² / 16) = 1 center: (0,6) a: 5 b: 4 c: 3

put in standard form, find center, find radius, and graph x² + y² + 2x + 8y + 8 = 0

standard form: (x + 1)² + (y + 4)² = -3 center: (-1,-4) radius: 3 quadrant it is in when graphed: third

put in standard form, find center, find radius, and graph x² + y² + 2x - 4y = 0

standard form: (x + 1)² + (y - 2)² = 5 center: (-1,2) radius: square root of 5 or approx. 2.2 quadrant it is in when graphed: second

put in standard form, find center, find radius, and graph x² + y² + 4x - 4y - 9 = 0

standard form: (x + 2)² + (y - 2)² = 17 center: (-2,2) radius: squared root of 17 or approx. 4.1 quadrant it is in when graphed: second

put in standard form, find center, find radius, and graph x² + y² + 4x - 8y + 19 = 0

standard form: (x + 2)² + (y - 4)² = 1 center: (-2,4) radius: 1 quadrant it is in when graphed: 2nd

put in standard form, find center, find radius, and graph x² + y² - 4x - 6y + 9 = 0

standard form: (x - 2)² + (y - 3)² = 4 center: (2,3) radius: 2 quadrant it is in when graphed: first

put in standard form, find center, find radius, and graph x² + y² - 2y - 8 = 0

standard form: x² + (y - 1)² = 9 center: (0,1) radius: 3 quadrant it is in when graphed: more in the northern hemisphere but relatively equal in all quadrants

the further the focus goes from the directrix,

the more the parabola widens (vice versa too)

ellipse

the set of all points in a plane such that the sum of the distances from two given points in the plane, called foci, is constant the sum of whose distances from 2 fixed points (foci) is constant

circle defintion

the set of points equidistant from a center point

write an expression for the distance between point P and the center of the circle

the square root of ((x - 3)² + (y - 5)²) = 4

write an expression for the distance between point P and the center of the circle

the square root of (x² + y²)

since you know the radius of the circle, you can turn your distance expression into an equation that describes all of the points in the circle. write and simplify the equation of this circle

the square root of (x² + y²) = 5 x² + y² = 25

horizontal lines would be

top and bottom

find transverse axis, a, b, c² = a² + b², c, asymptotes, and graph it (y² / 36) - (x² / 64) = 1

vertical (because y came first (first numerator)) 6 8 36 + 64 c = square root of 100 = 10 y = +/-(3/4)x center: (0,0) vertices: (0,6) (0,-6) foci: (0,10) (0,-10)

what would happen if the directrix were vertical instead of horizontal

vertical: side ways horizontal : up/down opening

what happenes as the focus moves further from the directrix?

wider parabola

directrix equation x vs y

x = is for if it is a vertical axis of symmetry(sides) y = is for if it is a horizontal axis of symmetry(up/down)

write equation focus (0,3) directrix y = -1

x² = 8(y - 1)

lookeat hyperbola notes for more information (good to look at) on the cone slicing

yah yah

review packet

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