9.1 Distance and Midpoint
9.2 Circles
9.3 Parabolas
9.4 Ellipses
9.5 Hyperbolas
9.6 More on Central Conics
9.7 Geometry of Quadratic Systems
9.8 Solving Quadratic Systems
9.9 Systems of Linear Equations with 3 Variables
100
Find the distance PQ.
P(-4,3) and Q(-6,-1)
PQ=root()((-6+4)^2+(-1-3)^2)=root()(4+16)=root()(20)=2root()(5)
100
Find an equation of the circle having center (4,-3) and radius 5.
(x-4)^2+(y+3)^2=25
100
Find an equation of the parabola having vertex (0,0) and directrix y=2.
c=2-0=2
a=1/(4c)=1/(4*2)=1/8
y-0=1/8(x-0)^2
y=1/8x^2
100
Find an equation of the ellipse having foci (0,-2) and (0,2) and sum of focal radii 8.
sum of focal radii = 8=2a
a=4
C(0,0)
c=2
a^2-c^2=b^2
16-4=12=b^2
b=root()(12)
(y^2)/16+(x^2)/12=1
100
pg. 452 #12
Find the foci of the hyperbola
8x^2-3y^2=48
x^2/6-y^2/16=1
a^2=6
b^2=16
c^2=a^2+b^2=6+16=22
c=root()(22)
(root()(22),0),(-root()(22),0)
100
pg 452 #14
Find the center of the conic
9x^2-y^2-18x+4y-31=0
(9x^2-18x+?)-(y^2-4y+?)=31
9(x^2-2x+1)-(y^2-4y+4)=31+9-4=36
9(x-1)^2-(y-2)^2=36
(x-1)^2/4-(y-2)^2/36=1
Hyperbola with center at (1, 2)
100
pg 452 #16
How many solutions does the system have?
x^2-4y^2=4
x+y^2=1
x^2-4y^2=4
x^2/4-y^2/1=1
-y^2=x-1
It is a hyperbola and a parabola
2 Solutions when you graph
100
pg 452 #17
Find the real solutions to the system
x^2+y^2=17
x^2-2y=9
x^2+y^2=17
-x^2+2y=-9
y^2+2y-8=0
(y+4)(y-2)=0
y=2, y=-4
x^2-2(2)=9
x^2=13
x=+-root()(13)
x^2-2(-4)=9
x^2=1
x=+-1
{(1,-4),(-1,-4),(root()(13),2),(-root()(13),2)}
100
pg 452 #19
Solve the system
x+y+z=2
x-2y-z=2
3x+2y+z=2
Add equations 1 and 2 you get
2x-y=4
Add equations 2 and 3 you get
4x=4
x=1
So now you have the three equations
x+y+z=2
2x-y=4
x=1
Solve from the bottom up
2(1)-y=4
y=-2
1-2+z=2
z=3
(1,-2,3)
200
Find the midpoint of PQ.
P(-4,3) and Q(-6,-1)
M=((-10)/2,2/2)=(-5,1)
200
Find the equation of the circle in the picture below.
(x+4)^2+(y-4)^2=25
200
Find the vertex of the parabola
4x=y^2-4y
4x=(y^2-4y+?)
4x+4=(y^2-4y+4)
4(x+1)=(y-2)^2
V(-1,2)
200
pg 452 # 11
Find the focus of the ellipse
x^2/4+y^2/20=1
a^2=20
b^2=4
a^2-b^2=c^2=20-4=16
c=4
F(0,-4) and (0,4)
200
pg 452 #13
Find the equation of the hyperbola at the right and has the vertices (2,0) and (-2,0). Is it
x^2-y^2=4
or
y^2-x^2=4
a/b=1/1
a=b
plug in (2,0)
4-0=4
0-4=4 no
200
pg 452 #15
Find an equation of the ellipse having foci (1,0) and (3,0) and sum of focal radii 4.
4=2a
a=2
Center is at (2, 0) and it is a horizontal ellipse
c=1
b^2=a^2-c^2=4-1=3
b=root()(3)
(x-2)^2/4+y^2/3=1
200
pg 453 #11
Sketch the graph and find out the number of solutions the system has.
4x^2+9y^2=36
x^2=y+2
x^2/9+y^2/4=1
x^2=y+2
Its an ellipse and a parabola. Looking at the graphs it has three solutions
200
pg452 #18
Find the dimensions of a rectangle having a perimeter of 14m and a diagonal of length 5m.
x+x+y+y=14
2x+2y=14
x+y=7
x^2+y^2=25
x=7-y
(7-y)^2+y^2=25
49-14y+y^2+y^2=25
2y^2-14y+24=0
y^2-7y+12=0
(y-4)(y-3)=0
y=3,4
x=7-3=4
x=7-4=3
rectangle is 3m by 4m
200
pg 453 #15
Solve the system
2x-y-3z=-1
2x-y+z=-9
x+2y-4z=17
Add the equation 2 to the opposite of equation 1 (multiply by -1) you get
4z=-8
z=-2
And add two of equation 2 to equation 3
5x-2z=-1
Now you have the three equations
2x-y-3z=-1
5x-2z=-1
z=-2
Solve from the bottom up
5x-2(-2)=-1
5x=-5
x=-1
2(-1)-y-3(-2)=-1
-y=-5
y=5
(-1,5,-2)
300
If Q is the midpoint of PM, find M.
P(-4,3) and Q(-6,-1)
(-6,-1)=((-4+x)/2,(3+y)/2)
2(-6)=(-4+x)
x=-8
2(-1)=(3+y)
y=-5
M(-8,-5)
300
Find the center and radius of the circle
x^2+y^2-8x+4y+12=0
(x^2-8x+?)+(y^2+4y+?)=-12
(x^2-8x+16)+(y^2+4y+4)=-12+16+4
(x-4)^2+(y+2)^2=8
h=(4,-2)
r=root()(8)=2root()(2)
300
Find an equation of the parabola shown below.
V(3,0)
It is negative and y is the squared variable because it is horizontal. So
x-3=-y^2
300
pg 422 #21
Find an equation of the ellipse having the given points as foci and the given number as the sum of focal radii.
(root()(5,0), (-root()(5),0);6
6=2a
a=3
c=root()(5)
a^2-c^2=b^2=9-5=4
b=2
x^2/9+y^2/4=1
300
pg 453 #8
Graph the hyperbola showing the asymptotes as dashed lines
x^2-y^2+4=0
Center=(0,0)
x^2-y^2=-4
y^2/4-x^2/4=1
a=2,a^2=4
b=2,b^2=4
c^2=a^2+b^2=4+4=8
c=root()(8)
asympt=b/ax=4/4x=x=y
asympt=-x=y
foci=(0,root()(8)),(0,-root()(8))
300
pg 453 # 10
Identify the conic and find its center and foci.
x^2-4y^2-4x-8y-4=0
(x^2-4x+?)-(4y^2+8y+?)=4
(x^2-4x+4)-4(y^2+4y+4)=4+4-16=-8
(x-2)^2-4(y+2)^2=-8
(y+2)^2/2-(x-2)^2/8=1
a^2=2
b^2=8
c^2=a^2+b^2=2+8=10
c=root()(10)
Hyperbola with center at (2,-2) and foci at
(2,-2+root()(10)), (2,-2-root()(10))
300
pg 453 #12
Sketch the graph and find the number of solutions in the system
4x^2-y^2=4
xy=-9
x^2/1-y^2/4=1
y=-9/x
It is a hyperbola and an inverse function. It has two solutions
300
pg 453 #14
Find the dimensions of a rectangle whose area is
36 m^2
and whose diagonal is
5root()(3)m
A=36=xy
x^2+y^2=(5root()(3))^2=75
x=36/y
(36/y)^2+y^2=75
1296/y^2+y^2-75=0
y^4-75y+1296=0
(y^2-27)(y^2-48)
y=root()(27)=3root()(3)
y=root()(48)=4root()(3)
dimensions are
4root()(3)m,3root()(3)m
300
pg 453 #20
Sara has $36 in $1, $5, and $10 bills, She has the same number of $5 bills as $10 bills, and she has 10 bills in all. How many bills of each denomination does she have?
x+5y+10z=36
x+y+z=10
y=z
Substitute equation 3 into 1 and 2
x+15y=36
x+2y=10
Add the opposite of the new equation 2 into the new equation 1
13y=26
y=2
z=2
x=6
(6,2,2)