9.1 Parabolas and Circles
9.2 Ellipses
9.3 Hyperbolas
9.6 Polar Coordinates
9.7 Polar Graphs
100
What is the equation of a circle with center at (3,4) and radius of 5?
(x-3)^2+(y-4)^2=25
100
Find the standard form of the equation of the ellipse and then graph it.
Vertices (2, 0), (2,4)
Foci: (2, 1), (2,3)
Center is halfway between vertices and foci.
C(2,2)
It is an up and down ellipse so major axis is on y
a is distance from center to vertices
a=2
c is the distance from center to foci
c=1
a^2-b^2=c^2
4-b^2=1
b=sqrt(3)
((y-2)^2)/4+((x-2)^2)/3=1
100
Find the standard form of the equation of the hyperbola and then graph it.
Vertices (2, 2), (-2,2)
Foci: (4,2), (-4,2)
The center is midway between the vertices so (0,2)
the distance from the center to the vertices is 2 so a=2
the distance from the center to the foci is 4 so c=4
a^2+b^2=c^2
4+b^2=16
b=sqrt(12)
(x^2)/4-((y-2)^2)/12=1
100
Plot the point given in polar coordinates and find the corresponding rectangular coordinates for the point.
(-4,2π/3)
x=rcos(theta)=-4cos((2π)/3)=2
y=rsin(theta)=-4sin((2π)/3)=-3.464
(2, -3.464)
100
Convert the polar equation into rectangular form.
r=10sintheta
r^2=10rsintheta
x^2+y^2=10y
x^2+(y^2-10y+25)=25
x^2+(y-5)^2=25
200
Find the standard form of the equation of the parabola. Then graph it.
Vertex (2, 2) Directrix: y=0
V(2,2) and D:y=0
p=2
Foci is (2,4)
Parabola has a vertical axis and is pointed up.
(x-h)^2=4p(y-k)
(x-2)^2=4(2)(y-2)
(x-2)^2=8(y-2)
200
A semielliptical archway is to be formed over the entrance to an estate. The arch is to be set on pillars that are 10 feet apart and is to have a height (atop the pillars of 4 feet. Where should the foci be placed to sketch the arch?
pillars 10 feet apart mean that the vertices are 10 feet apart so a=5. Half of the minor axis is 4 feet high so b=4.
a^2-b^2=c^2
25-16=9
c=3
the foci will be 2 feet from the pillars or at (3,0) and (-3,0) (assuming the center is at (0,0) and the pillars are at (5,0) and (-5,0))
200
Find the (a)standard form of the equation of the ellipse,(b)find the center, vertices, foci, and eccentricity of the hyperbola, and (c) sketch the ellipse.
-4x^2+25y^2-8x+150y+121=0
(-4x^2-8y+?)+(25y^2+150y+?)=-121
-4(x^2+2y+1)+25(y^2+6y+9)=-121-4+225
-4(x+1)^2+25(y+3)^2=100
((y+3)^2)/4-((x+1)^2)/25=1
4+25=c^2
c=sqrt(29)
Center (-1,-3)
Vertices (-1, -1),(-1,-3)
Foci (-1,3+sqrt(29)),(-1,3-sqrt(29))
e=sqrt(29)/4
200
Given the rectangular coordinate, find an equivalent polar coordinate.
(-3,4)
tan(theta)=(4/-3)
theta=-.927"rad"=-53.13°
x^2+y^2=r^2=9+16=25
r=5
(5, .927)
200
Sketch the graph of (include maximums and zeros)
r=3-5sin(theta)
Maximum at (8,3π/2)
Zeros at (0,0.644) and (0,0.249)
300
The light bulb in a flashlight is at the focus of its parabolic reflector, 1.5 cm from the vertex of the reflector. Write an equation for the cross-section of the flashlight's reflector with its focus on the positive x-axis and its vertex on the origin.
p=1.5, V(0,0) and its going to be a horizontal parabola pointed in the positive x axis.
(y-0)^2=4(1.5)(x-0)
y^2=6x
300
Find the (a)standard form of the equation of the ellipse,(b)find the center, vertices, foci, and eccentricity of the ellipse, and (c) sketch the ellipse.
4x^2+25y^2+16x-150y+141=0
4x^2+25y^2+16x-150y+141=0
4(x^2+4x+?)+25(y^2-6y+?)=-141
4(x^2+4x+4)+25(y^2-6y+9)=-141+16+225
4(x+2)^2+25(y-3)^2=100
((x+2)^2)/25+((y-3)^2)/4=1
25-4=c^2
c=sqrt(21)
Center(-2,3)
Vertices(3,3),(-7,3)
e=sqrt(21)/5
Foci (-2+sqrt(21),3)(-2-sqrt(21),3)
300
Classify the graph of the equation as a circle, parabola, an ellipse or a hyperbola.
3x^2+2y^2-12x+12y+29=0
(3x^2-12x+?)+(2y^2+12y+?)=-29
3(x^2-4x+4)+2(y^2+6y+9)=-29+12+18
3(x-2)^2+2(y+3)^2=1
((x-2)^2)/(1/3)+((y+3)^2)/(1/2)=1
Ellipses
300
Convert the rectangular equation into polar form.
xy=-2
rcos(theta)rsin(theta)=-2
r=-2/(cos(theta)sin(theta))
300
Sketch the graph of (include maximums and zeros)
r=-3cos(2theta)
maximums at (3,0),(3,π/2),(3,π),(3,3π/2),(3,2π)
zeros at (0,π/4)(0,3π/4)(0,5π/4)(0,7π/4)
