unit 3
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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