Conics Packet
Honors Geometry
HW #1
For each parabola, find the vertex, focus, and equation of the directrix. Then
draw the graph (without a calculator). You should check the graph with your
calculator.
1. y 2  6 x
2. y 2  4 x  0
Find an equation of a parabola satisfying the given conditions.
3. Focus (4,0) and directrix x  4
4. Focus ( 2 ,0) and directrix x  2
For each parabola, find the vertex, focus, and equation of the directrix. Then
draw the graph (without a calculator). You should check the graph with your
calculator.
5. ( x  2) 2  6( y  1)
6. y 2  6 y  x  16  0
7. y  x 2  4 x  3
8. y 2  y  x  4  0
9. Assume that point P with coordinate (x,y) is equidistant to the line y=2 and the point (3,4).
a.
In terms of x and y, the distance from P to the line y=2 is __________.
b. The distance from P to the point (3,4) is ______________.
10. Set the two distances equal from #9, simplify (by squaring both sides first), and isolate the y
term.
11. On a sheet of graph paper, plot points to graph the equation that you found in #11. What
kind of equation do you have?
Extra practice: text, 10.1 (p. 820-22) # 7-30
Answers:
1.
2.
3.
4.
5.
6.
7.
8.
v=(0,0); F=(-3/2,0); D:x=3/2
v=(0,0); F=(-1,0); D:x=1
y 2  16 x
y 2  4 2 x
v=(-2,1); F=(-2,-1/2); D:y=5/2
v=(7,-3); F=(29/4,-3); D:x=27/4
v=(-2,-1); F=(-2,-3/4); D:x=-5/4
v=(-17/4,-1/2); F=(-4,-1/2); D:x=-9/2
| y-2 |; ( x  3) 2  ( y  4) 2
1
3
21
10. y  x 2  x 
4
2
4
11. parabola
9.
HW #2:
Correct homework from previous night AND do:
1. Find an equation of a parabola with focus (3,2) and directrix x = –4
2. Find the vertex, focus and directrix of the parabola with equation
x 2  2x  2 y  7  0
3. An engineer designs a satellite dish with a [parabolic cross section.
The dish is 15 feet wide at the opening and the focus is placed 4 feet
from the vertex.
15 ft
4 ft
focus
a. Position a coordinate system with the origin at the vertyex and x-axis
on the parabola’s axis of symmetry and find an equation of the
parabola.
b. Find the depth of the satellite dish.
4. A headlight mirror has a parabolic cross section
with diameter 6 inches and depth 1 inch.
a. Position a coordinate system with the origin at the
vertex and the x-axis on the parabola’s axis of
symmetry, and find an equation of the parabola.
b. How far from the vertex should the bulb be
positioned if it is to be placed at the focus?
6 in.
1 in.
focus
5. Find the center and radius of the circle x 2  y 2  6 x  2 y  6
6. Find the equation of the parabola which is the set of points equidistant to the point (3,-4)
and the line y=-1. Do this by writing an expression for the distance from the point (x,y) to
the the point (3,-4) and an expression for the distance from (x,y) to the line y=-1. Set the
distances equal and simplify.
Extra practice: text, 10.1 (p. 820-22) # 7-30, supplement #1
Answers:
1.
1
( y  2) 2  14( x  )
2
2.
V: (-1,-3); F: (-1,-7/2); D:y=5/2
y 2  16 x
3.
4.
3
33
64
y 2  9x
2.25
5.
ctr: (-3,1); radius=4
1
6. y   x 2  x  4
6
HW#3
1. Find the center and radius of the circle x 2  y 2  6 x  8 y  16  0
Find the vertices and foci of each ellipse. Graph, showing the endpoints
of the minor axis.
2.
x2 y2

1
25 36
3. 2 x 2  3 y 2  6
4. 25 x 2  16 y 2  1
5. Find the equation of an ellipse with vertices (0,-8) and (0,8) and the
length of the minor axis = 10.
6. Find the equation of an ellipse with foci (0,-3) and (0,3) and the length
of the major axis = 10.
Find the center, the vertices, and the foci of each ellipse. Graph, showing
the length of the minor axis.
7.
( x  2) 2 ( y  3) 2

1
16
25
8. 4( x  5) 2  3( y  4) 2  48
9. 4 x 2  y 2  24 x  2 y  21  0
Extra practice: section 10.2 (p 830-32)
Answers:
1. ctr (3,4) rad = 3
2. V (0,6),(0,-6); F: (0, 11 ), (0,- 11 ); minor axes ends: (5,0),(-5,0)
3. V (- 3 ,0),( 3 ,0); F: (-1,0), (1,0); minor axes ends: (0, 2 ),(0,- 2 )
4. V (0,-1/4),(0,1/4); F: (0,-3/20), (0,3/20);
minor axes ends: (1/5,0),(-1/5,0)
x2

25
x2
6.

16
5.
y2
1
64
y2
1
25
7. C(2,-3); V(2,-8), (2,2); F(2,-6),(2,0); len of minor axis = 8
8. C(5,4); V(5,0), (5,8); F(5,2),(5,6); ends minor axis = (5  2 3,4)
9. Ctr=(-3,1), F: (3,1  2 3 ) V(-3,5),(-3,-3), ends minor axis: (-5,1),(-1,1)
HW #4
1. Sketch the graph of each "half conic" (include at least 2 points).
a.
y= x
d. y 
b.
y = - x-2 +3
c. x = - 9 - y 2
4  x2
2
2. A satellite dish is shaped like a paraboloid. If the dish is 24
inches across and 4 inches deep, where should the
transmission collector be positioned?
3. Describe the set of points satisfying each equation
(x
2
)(
)
- y y2 - x = 0
4. Find an equation of each ellipse described:
a. Vertices are 5,9 and 5,1 and one focus is 5,7 .
b. Foci are at 0,3 and 10,3 and its major axis is 26 units long.
5.
6.
A hall 100 feet in length is to be designed as a whispering gallery. If the foci are
located 25 feet from the center, how high will the ceiling be at the center?
If A=(0,2) and B=(0,-2), E=(x,y), and AE + EB = 12, express this fact in an equation in terms
of x and y.
Simplify the equation. You will need to square both sides of the equation twice. Remember,
when you square a binomial, you get a trinomial, so do not forget the middle term! (The first
time, put one radical on each side of the equation before squaring.)
x2 y2
Complete the derivation to write in the form 2 + 2 = 1
a
b
Graph your equation by solving for y and using your graphing calculator. (hint: Don’t
forget the +/- when dealing with squares)
7.
8.
For an ellipse with center at (0,0), let a focus be at (c,0) and an intercept at (a,0). The
quotient c/a is the eccentricity of the ellipse. Thinking about creating an ellipse with string,
a.
What is the largest possible eccentricity?
b.
What is the smallest eccentricity?
c.
How is the shape different for a small eccentricity vs. a large one?
Find an equation of an ellipse with vertices (0,-4) and (0,4) and eccentricity =
Extra practice: section 10.2 (p 830-32), Conics Supplement #1
Answers:
1. a. y = x
b. y = - x - 2 + 3
1
4
c. left half of a circle with radius 3 centered at the origin.
d. Top half of an ellipse through (2,0),(-2,0), (0,1)
2.
The collector should be 9 inches in front of the vertex of the paraboloid.
3.
Two parabola with equations y  x 2 and x  y 2
4.
(x - 5)2 + (y - 5)2
=1
12
16
(x - 5)2 + (y + 3)2 = 1
b.
169
144
a.
5. 43.3 feet tall
6.
7.
8.
x2 y2

1
32 36
a. approaches 1 b. approaches 0 c. closer to zero: more circular; closer to 1:
flatter
x2 y2

1
15 16
HW#5
Find the center, vertices, and equations of asymptotes, and sketch a graph:
1.
x2 y2

1
1
9
2.
( y  3) 2 ( x  1) 2

1
1
9
3. 4 x 2  y 2  16
4. x 2  y 2  2
5. y 2  x 2 
1
9
6. 36 x 2  y 2  24 x  6 y  41  0
7. Point A is (-3,0) and point B is (3,0). Let point E be (x,y). Write an equation
to express the fact that | AE – EB | = 2. Simplify.
5
4
5
4
8. Find an equation of a hyperbola with asymptotes y  x and y   x .
and one vertex is (0,3)
Additional practice: sec 10-3 (p 840-42)
Answers:
(answers include the foci, which you will be doing in a future
assignment)
1. C(0,0); V: (-1,0),(1,0); F(- 10 ,0),( 10 ,0); A: y=3x, y=-3x
2. C(-1,-3); V: (-1,-2),(-1,-4); F(-1,-3+ 10 ),(-1,-3- 10 ); A:
y+3=1/3(x+1). y+3=-(1/3)(x+1)
3. C(0,0); V: (-2,0),(2,0); F(-2 5 ,0),(2 5 ,0); A: y=2x. y=-2x
4. C(0,0); V: (- 2 ,0),( 2 ,0); F(2,0),(-2,0); A: y=x. y=-x
2
2
), (0,- ) A: y=x. y=-x
3
3
1
1
6. C(1/3,3); V: (-2/3,3),(4/3,3); F(  37 ,3), (  37 ,3), A: y=6x+1.
3
3
5. C(0,0); V: (0,-1/3),(0,1/3); F(0,
y=-6x+5
7.
x2 y2

1
1
8
8. -
x2
y2
+
=1
2
92
æ 12 ö
çè ÷ø
5
HW#6
1. Find the foci from HW#5, problems #2 through 4
2
3
2
3
2. Find an equation of a hyperbola with asymptotes y  x and y   x
and one vertex is (6,0)
3. Sketch xy  12
4. Write an equation for each hyperbola described:
a. Center is 5,0 , one vertex is 9,0 and one focus is 10,0
b. Vertices are 4,0 and 4,20 and the eq. of one asymptote is y 
2
42
x
5
5
For #6 and 7, sketch a graph of each of the following conic sections. Be sure to find the
following:
For ellipses: center, vertex points, and foci
For hyperbolas: center, vertex points, foci, and the equations for the asymptotes
For parabolas: vertex, focus, and the equation for the directrix
5. 5x 2  16 y 2  30 x  416 y  2579  0
6. 3 y 2  8x  12 y  52  0
7. The ellipses 4 x 2  5 y 2  81 and 5x 2  4 y 2  81 intersect.
a. Where is/are the point(s) of intersection? How many will there be? Sketch the
intersection(s).
b. Add the equation of the two ellipses together, what relationship does this new
equation have to the points of intersection?
c. Why is this relationship true? Make an educated hypothesis.
Additional practice: Conics Supplement #2
Answers:
1. see assignment #5 answers
x2 y2

1
36 16
3. xy  12
2.
4. a.
 x  5 2
16

y2
1
9
b. 
5. Hyperbola: 
x  42   y  102
625
x  32   y  132
16
Center:  3,13

Vertices:  3,13  5

100
Foci:  3,13  21

5
1
1

Asymptotes: y  13  
5
 x  3
4
3
 y  2 2  x  5
8
Opens to the left
Vertex:  5,2
 17

Focus:   ,2 
 3

6. Parabola: 
7. a. the two ellipses have 4 points of intersection  3,3
b. adding the two equations together creates the equation for the circle with center at
the origin and a radius of 3 2 . The circle contains the 4 points of intersection.
c. Answers will vary. Any combination of equations within a system creates another
equation with that contains the points of intersection.
HW#7
Watch this: http://youtu.be/v-pyuaThp-c
1. Solve each system of equations.
a.
x 2 + y 2 = 25
y- x =1
b.
x 2 + 4y 2 = 20
xy = 4
Sketch a graph of each of the following conic sections. Be sure to find the following:
For ellipses: center, vertex points, and foci
For hyperbolas: center, vertex points, foci, and the oblique (slant) asymptotes
For parabolas: vertex, focus, and the equation for the directrix
2. 24 x 2 + 16 y 2 + 240 x - 96 y + 552 = 0
3. 2 x 2 + 16 x + 12 y - 28 = 0
4. 16 x 2 - 9 y 2 + 160 x + 18 y + 535 = 0
5. 4 x 2 - 7 y 2 - 32 x - 98 y - 307 = 0
6. Find a polynomial equation for each locus described. You do not need to simplify. If
the equation is that of a conic section, tell which conic section it is.
a. The locus of points whose distance from (-3,0) is twice its distance from (3,0).
b. The locus of points whose distance from y  1 is one-third its distance from
x  2.
c. The locus of points whose distance from (-3,1) is half its distance from y  4 .
7. Find a polynomial equation of the locus of points whose distance from 0, f  is e times
its distance from y   f (where e is a constant). Only simplify enough to show the
following: Show that the graph will be an ellipse if 0  e  1 ; a parabola if e  1; and a
hyperbola if e  1.
Optional review: p 883-84#1-20, or do extra problems from any of the
Supplements #1, 2 or 3.
Answers
1. a. (3,4) and (-4,-3) b. (4,1), (-4,-1), (2,2) and (-2,-2) (Note: this is not the
same thing as ( ±2, ±2 ) and ( ±4, ±1) ).
2. Ellipse:
(x + 5)2 + (y - 3)2
8
12
Center: (- 5,3)
(
=1
)
Vertices: - 5,3 ± 2 3
Foci: (- 5,5) and (- 5,1)
1
2
3. Parabola: y - 5 = - (x + 4)
6
Vertex: (- 4,5)
7ö
æ
Focus: ç - 4, ÷
2ø
è
13
Directrix: y =
2
(x + 5)2 + (y - 1)2 = 1
4. Hyperbola: 9
16
Center: (- 5,1)
Vertices: (- 5,5) and (- 5 - 3)
Foci: (- 5,6) and (- 5 - 4)
4
Asymptotes: (y - 1) = ± (x + 5)
3
5. Hyperbola:
(x - 4)2 - (y + 7)2
Center: (4,-7 )
7
(
4
Vertices: 4 ± 7 ,-7
(
Foci: 4 ± 11,-7
)
=1
)
Asymptotes: (y + 7 ) = ±
2
7
(x - 4)
6. a. (x + 3)2 + y 2 = 2 (x - 3)2 + y 2 - x 2 + y 2 - 10 x + 9 = 0 - a circle!!
more specifically: (x - 5) + y 2 = 16
2
b. y - 1 =
1
1
x - 2 or y - 1 = ± (x - 2) - two intersecting lines (a degenerate
3
3
hyperbola)
c.
(x + 3)2 + (y - 1)2
=
1
y-4
2
>>>
2 (x + 3) + (y - 1) = y - 4
2
2
4 x 2 + 3 y 2 + 24 x + 24 = 0 - an ellipse
7.
x 2 + (y - f ) = e y + f
2
x 2 + y 2 - 2 fy + f 2 = e 2 y 2 + 2e 2 fy + e 2 f 2






x 2  1  e 2 y 2  2 fy 1  e 2  f 2 1  e 2  0
If
If
If
If
e = 0 , quadratic term coefficients are equal so it’s a circle.
0 < e < 1, quadratic term coefficients are both positive so it’s still a circle!
e = 1, the y 2 term goes away so it’s a parabola.
e > 1, the quadratic term coefficients will have opposite signs so it’s a
hyperbola.