19 more parabolas a& hyperbolas (optional) x

Hyperbolas and More Parabolas (optional)

Using similar methods to analyze Ax2 + By2 + Cx + Dy = E
we obtain the graphs of hyperbolas and parabolas.

In the special case B = 0 (or A = 0),

In the special case B = 0 (or A = 0), after dividing by A (by B),
the equation becomes 1x2 + #x + #y = # or 1y2 + #x + #y = #,

their graphs are parabolas. Assuming both variables x and y
remained in the equation
(
(

their graphs are parabolas.
1x2 + #x + #y = # or
1y2 + #x + #y = #
Parabolas:
Assuming both variables x and y
(
(

1x2 + #x + #y = # or
1y2 + #x + #y = #
If the equation Ax2 + By2 + Cx + Dy = E
has A and B of opposite signs,
Parabolas:
(
(

1x2 + #x + #y = # or
1y2 + #x + #y = #
has A and B of opposite signs, after dividing by A, we have 1x2
+ ry2 + #x + #y = #, with r < 0.
Parabolas:
(
(

1x2 + #x + #y = # or
1y2 + #x + #y = #
+ ry2 + #x + #y = #, with r < 0. These are hyperbolas.
Parabolas:
(
(

1x2 + #x + #y = # or
1y2 + #x + #y = #
(r < 0)
Hyperbolas: 1x2 + ry2 + #x + #y = #
+ ry2 + #x + #y = #, with r < 0. These are hyperbolas.
Parabolas:
their graphs are parabolas. Assuming both variables x and y
(
(
(in general)

Conic Sections
After dividing Ax2 + By2 + Cx + Dy = E by A, we obtain
1x2 + ry2 + #x + #y = #.

Conic Sections
1x2 + ry2 + #x + #y = #. Let’s use the simpler cases 1x2 + ry2 = 1
to study the geometry.

Conic Sections
to study the geometry. Starting with r = 1, a circle,
circle
r = 1

Conic Sections
as r becomes smaller = ¼,
circle
r = 1

Conic Sections
as r becomes smaller = ¼, 1/9, 1/16… ,
circle
r = 1

Conic Sections
1x2 + y2 = 1
1
16
4
1
circle
r = 1

Conic Sections
the circle is elongated to taller and taller ellipses.
circle ellipses
1x2 + y2 = 1
1
16
4
1
r = 1

Conic Sections
When r = 0, we’ve 1x2 + 0y2 = 1,
circle ellipses
. … r → 0
1x2 + y2 = 1
1
16
r = 1/16
4
1
r = 1

Conic Sections
When r = 0, we’ve 1x2 + 0y2 = 1, which is a pair of lines x = ±1,
circle ellipses
. … r → 0
1x2 + y2 = 1
1
16
r = 1/16
4
1
r = 1

Conic Sections
circle ellipses
. … r → 0
1x2 + y2 = 1
1
16
r = 1/16
4
1
1
1x2 = 1
or
x = ±1
r = 1 r = 0
two lines

Conic Sections
i.e. the ellipses are elongated
into two parallel lines.
circle ellipses
. … r → 0
1x2 + y2 = 1
1
16
r = 1/16
4
1
1
1x2 = 1
or
x = ±1
r = 1 r = 0
two lines

Hyperbolas
Just as all the other conic sections, hyperbolas are defined
by distance relations.

Hyperbolas
Given two fixed points, called foci, a hyperbola is the set
of points whose difference of the distances to the foci is
a constant.

A
If A, B and C are points on a hyperbola as shown
Hyperbolas
a constant.
B
C

A
a2
a1
If A, B and C are points on a hyperbola as shown then
a1 – a2
Hyperbolas
a constant.
B
C

A
a2
a1
b2
b1
a1 – a2 = b1 – b2
Hyperbolas
a constant.
B
C

A
a2
a1
b2
b1
a1 – a2 = b1 – b2 = c2 – c1 = constant.
c1
c2
Hyperbolas
a constant.
B
C

Hyperbolas
A hyperbola has a “center”,

Hyperbolas
A hyperbola has a “center”, and two straight lines that
cradle the hyperbolas which are called asymptotes.

Hyperbolas
There are two vertices, one for each branch.

Hyperbolas
There are two vertices, one for each branch. The asymptotes
are the diagonals of a rectangle with the vertices of the
hyperbola touching the rectangle.

Hyperbolas
The center-rectangle is defined by the x-radius a, and y-
radius b as shown.
a
b

Hyperbolas
radius b as shown. Hence, to graph a hyperbola, we find
the center and the center-rectangle first.
a
b

Hyperbolas
the center and the center-rectangle first. Draw the
diagonals of the rectangle which are the asymptotes.
a
b

Hyperbolas
diagonals of the rectangle which are the asymptotes. Label
the vertices and trace the hyperbola along the asymptotes.
a
b

Hyperbolas
a
b
The location of the center, the x-radius a, and y-radius b may
be obtained from the equation.
diagonals of the rectangle which are the asymptotes. Label
the vertices and trace the hyperbola along the asymptotes.

Hyperbolas
The equations of hyperbolas have the form
Ax2 + By2 + Cx + Dy = E
where A and B are opposite signs.

Hyperbolas
Ax2 + By2 + Cx + Dy = E
where A and B are opposite signs. By completing the square,
they may be transformed into the standard forms below.

(x – h)2 (y – k)2
a2 b2
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2 – = 1

(x – h)2 (y – k)2
a2 b2
(h, k) is the center.
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2 – = 1

(x – h)2 (y – k)2
a2 b2
x-rad = a, y-rad = b
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2 – = 1

(x – h)2 (y – k)2
a2 b2
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2
y-rad = b, x-rad = a
– = 1

(x – h)2 (y – k)2
a2 b2
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2
– = 1
(h, k)
Open in the x direction

(x – h)2 (y – k)2
a2 b2
Hyperbolas
Ax2 + By2 + Cx + Dy = E
– = 1 (x – h)2
(y – k)2
a2
b2
– = 1
(h, k)
Open in the x direction
(h, k)
Open in the y direction

Hyperbolas
Following are the steps for graphing a hyperbola.

Hyperbolas
1. Put the equation into the standard form.

Hyperbolas
2. Read off the center, the x-radius a, the y-radius b, and
draw the center-rectangle.

Hyperbolas
3. Draw the diagonals of the rectangle, which are the
asymptotes.

Hyperbolas
asymptotes.
4. Determine the direction of the hyperbolas and label the
vertices of the hyperbola.

Hyperbolas
asymptotes.
vertices of the hyperbola. The vertices are the mid-points
of the edges of the center-rectangle.

Hyperbolas
asymptotes.
vertices of the hyperbola. The vertices are the mid-points
of the edges of the center-rectangle.
5. Trace the hyperbola along the asymptotes.

Hyperbolas
Example A. List the center, the x-radius, the y-radius.
Draw the rectangle, the asymptotes, and label the vertices.
Trace the hyperbola.
(x – 3)2 (y + 1)2
42 22
– = 1

Center: (3, -1)
Hyperbolas
(x – 3)2 (y + 1)2
42 22
– = 1

Center: (3, -1)
x-rad = 4
y-rad = 2
Hyperbolas
(x – 3)2 (y + 1)2
42 22
– = 1

Center: (3, -1)
x-rad = 4
y-rad = 2
Hyperbolas
(3, -1)
4
2
(x – 3)2 (y + 1)2
42 22
– = 1

Center: (3, -1)
x-rad = 4
y-rad = 2
The hyperbola opens
left-rt
Hyperbolas
(x – 3)2 (y + 1)2
42 22
– = 1
(3, -1)
4
2

Center: (3, -1)
x-rad = 4
y-rad = 2
The hyperbola opens
left-rt and the vertices
are (7, -1), (-1, -1) .
Hyperbolas
(x – 3)2 (y + 1)2
42 22
– = 1
(3, -1)
4
2

Center: (3, -1)
x-rad = 4
y-rad = 2
The hyperbola opens
are (7, -1), (-1, -1) .
Hyperbolas
(3, -1)
(7, -1)
(-1, -1) 4
2
(x – 3)2 (y + 1)2
42 22
– = 1

Center: (3, -1)
x-rad = 4
y-rad = 2
The hyperbola opens
are (7, -1), (-1, -1) .
Hyperbolas
(3, -1)
(7, -1)
(-1, -1) 4
2
(x – 3)2 (y + 1)2
42 22
– = 1
When we use completing the square to get to the standard
form of the hyperbolas, depending on the signs,
we add a number or subtract a number from both sides.

Example B. Put 4y2 – 9x2 – 18x – 16y = 29 into the standard
form. List the center, the x-radius, the y-radius.
Hyperbolas

Group the x’s and the y’s:
Hyperbolas

4y2 – 16y – 9x2 – 18x = 29
Hyperbolas

4y2 – 16y – 9x2 – 18x = 29 factor out the square-coefficients
Hyperbolas

4(y2 – 4y ) – 9(x2 + 2x ) = 29
Hyperbolas

4(y2 – 4y ) – 9(x2 + 2x ) = 29 complete the square
Hyperbolas

4(y2 – 4y + 4 ) – 9(x2 + 2x ) = 29
Hyperbolas

4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29
Hyperbolas

4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16
16
Hyperbolas

4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas

4(y – 2)2 – 9(x + 1)2 = 36 divide by 36 to get 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas

9(x + 1)2
4(y – 2)2
36 36
– = 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas

9(x + 1)2
4(y – 2)2
36 36
– = 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas
9

9(x + 1)2
4(y – 2)2
36 36
– = 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas
9 4

9(x + 1)2
4(y – 2)2
36 36
– = 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas
(y – 2)2 (x + 1)2
32 22
– = 1
9 4

9(x + 1)2
4(y – 2)2
36 36
– = 1
4(y2 – 4y + 4 ) – 9(x2 + 2x + 1 ) = 29 + 16 – 9
16 –9
Hyperbolas
(y – 2)2 (x + 1)2
32 22
– = 1
Center: (-1, 2), x-rad = 2, y-rad = 3
The hyperbola opens up and down.
9 4

(-1, 2)
Hyperbolas
Center: (-1, 2),
x-rad = 2,
y-rad = 3

(-1, 2)
(-1, 5)
(-1, -1)
Hyperbolas
Center: (-1, 2),
x-rad = 2,
y-rad = 3
The hyperbola opens up and down.
The vertices are (-1, -1) and (-1, 5).

The graphs of the equations of the form
y = ax2 + bx + c and x = ay2 + bx + c
are parabolas.
More Graphs of Parabolas

Each parabola has a vertex and the center line that contains
the vertex.
are parabolas.

the vertex. Suppose we know another point on the parabola,
are parabolas.

then the reflection of the point across the center line is also
on the parabola.
are parabolas.

then the reflection of the point across the center line is also
on the parabola. There is exactly one parabola that goes
through these three points.
are parabolas.

When graphing parabolas, we must also give the x-intercepts
and the y-intercept.

The y-intercept is obtained by setting x = 0 and solve for y.

The x-intercept is obtained by setting y = 0 and solve for x.

The graphs of y = ax2 + bx = c are up-down parabolas.

If a > 0, the parabola opens up.
a > 0

If a < 0, the parabola opens down.
a > 0 a < 0

a > 0 a < 0
Vertex Formula (up-down parabolas) The x-coordinate of
the vertex of the parabola y = ax2 + bx + c is at x = .
-b
2a

Following are the steps to graph the parabola y = ax2 + bx + c.

1. Set x = in the equation to find the vertex.
-b
2a

2. Find another point, use the y-intercept (0, c) if possible.
-b
2a

3. Locate its reflection across the center line, these three
points form the tip of the parabola. Trace the parabola.
-b
2a

4. Set y = 0 and solve to find the x intercept.
-b
2a

Example A. Graph y = –x2 + 2x + 15
-b
2a

The vertex is at x = = 1
–(2)
2(–1)
-b
2a

The vertex is at x = = 1y = 16.
–(2)
2(–1)
-b
2a

Example A. Graph y = –x2 + 2x + 15 (1, 16)
–(2)
2(–1)
-b
2a

The y-intercept is at (0, 15)
–(2)
2(–1)
-b
2a
(1, 16)

–(2)
2(–1)
-b
2a
(1, 16)
(0, 15)

Plot its reflection (2, 15).
–(2)
2(–1)
-b
2a
(1, 16)
(0, 15)

Plot its reflection (2, 15).
–(2)
2(–1)
-b
2a
(1, 16)
(0, 15) (2, 15)

Plot its reflection (2, 15)
Draw,
(0, 15) (2, 15)
–(2)
2(–1)
-b
2a

Draw, set y = 0 to get x-int:
(0, 15) (2, 15)
–(2)
2(–1)
-b
2a

–x2 + 2x + 15 = 0
(0, 15) (2, 15)
–(2)
2(–1)
-b
2a

–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(0, 15) (2, 15)
–(2)
2(–1)
-b
2a

–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5
(0, 15) (2, 15)
–(2)
2(–1)
-b
2a

–x2 + 2x + 15 = 0
x2 – 2x – 15 = 0
(x + 3)(x – 5) = 0
x = –3, x = 5
(0, 15) (2, 15)
(-3, 0) (5, 0)
–(2)
2(–1)
-b
2a

The graphs of the equations
x = ay2 + by + c
are parabolas that open sideways.

x = ay2 + by + c
If a>0, the parabola opens
to the right.

x = ay2 + by + c
to the right.
If a<0, the parabola opens
to the left.

x = ay2 + by + c
Each sideways parabola is symmetric to a horizontal center
line.
to the right.
to the left.

x = ay2 + by + c
line. The vertex of the parabola is on this line.
to the right.
to the left.

x = ay2 + by + c
line. The vertex of the parabola is on this line. If we know the
location of the vertex and another point on the parabola, the
parabola is completely determined.
to the right.
to the left.

x = ay2 + by + c
line. The vertex of the parabola is on this line. If we know the
location of the vertex and another point on the parabola, the
parabola is completely determined. The vertex formula is the
same as before except it's for the y coordinate.
to the right.
to the left.

Vertex Formula (sideways parabolas)
The y coordinate of the vertex of the parabola
x = ay2 + by + c
is at y = .
–b
2a

Following are steps to graph the parabola x = ay2 + by + c.
x = ay2 + by + c
is at y = .
–b
2a

1. Set y = in the equation to find the x coordinate of the
vertex.
–b
2a
x = ay2 + by + c
is at y = .
–b
2a

vertex.
2. Find another point; use the x intercept (c, 0) if it's not the
vertex.
–b
2a
x = ay2 + by + c
is at y = .
–b
2a

vertex.
vertex.
3. Locate the reflection of the point across the horizontal
center line, these three points form the tip of the parabola.
Trace the parabola.
–b
2a
x = ay2 + by + c
is at y = .
–b
2a

vertex.
vertex.
3. Locate the reflection of the point across the horizontal
center line, these three points form the tip of the parabola.
Trace the parabola.
4. Set x = 0 to find the y intercept.
–b
2a
x = ay2 + by + c
is at y = .
–b
2a

Example B. Graph x = –y2 + 2y + 15

Vertex: set y = =1
–(2)
2(–1)

Vertex: set y = =1 then x = –(1)2 + 2(1) + 15 = 16
–(2)
2(–1)

–(2)
2(–1)
so v = (16, 1).

–(2)
2(–1)
so v = (16, 1).
Another point:
Set y = 0 then x = 15
or (15, 0).

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
(15, 0)
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
Plot its reflection.
(15, 0)
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2).
(15, 0)
(16, 1)
(15, 2)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2).
Draw. (15, 0)
(16, 1)
(15, 2)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2)
Draw. (15, 0)
(15, 2)
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2)
Draw. Get y-int:
–y2 + 2y + 15 = 0
(15, 0)
(15, 2)
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2)
Draw. Get y-int:
–y2 + 2y + 15 = 0
y2 – 2y – 15 = 0
(y – 5) (y + 3) = 0
y = 5, -3
(15, 0)
(15, 2)
(16, 1)

–(2)
2(–1)
so v = (16, 1).
Another point:
or (15, 0).
It's (15, 2)
Draw. Get y-int:
–y2 + 2y + 15 = 0
y2 – 2y – 15 = 0
(y – 5) (y + 3) = 0
y = 5, -3
(15, 0)
(15, 2)
(16, 1)
(0, -3)

The y-intercept is (o, c) obtained by setting x = 0.
The x-intercept is obtained by setting y = 0 and solve the
equation 0 = ax2 + bx + c which may or may not have real
number solutions. Hence there might not be any x-intercept.
Following are the steps to graph a parabola y = ax2 + bx + c.
-b
2a
The graph of y = ax2 + bx = c are up-down parabolas.

Exercise A. Graph the following parabolas. Identify which
direction the parabolas face, determine the vertices using
the vertex method. Label the x and y intercepts, if any.
1. x = –y2 – 2y + 15 2. y = x2 – 2x – 15
3. y = x2 + 2x – 15 4. x = –y2 + 2y + 15
5. x = –y2 – 4y + 12 6. y = x2 – 4x – 21
7. y = x2 + 4x – 12 8. x = –y2 + 4y + 21
9. x = –y2 + 4y – 4 10. y = x2 – 4x + 4
11. x = –y2 + 4y – 4 12. y = x2 – 4x + 4
13. y = –x2 – 4x – 4 14. x = –y2 – 4y – 4
15. x = –y2 + 6y – 40 16. y = x2 – 6x – 40
17. y = –x2 – 8x + 48 18. x = y2 – 8y – 48
19. x = –y2 + 4y – 10 20. y = x2 – 4x – 2
21. y = –x2 – 4x – 8 22. x = –y2 – 4y – 5

Ellipses
B. Complete the square of the following hyperbola-equations
if needed. Find the centers and the radii of the hyperbolas.
Draw and label the vertices.
1. x2 – 4y2 = 1 2. 9x2 – 4y2 = 1
3. 4x2 – y2/9 = 1 4. x2/4 – y2/9 = 1
5. 0.04x2 – 0.09y2 = 1 6. 2.25x2 – 0.25y2 = 1
7. x2 – 4y2 = 100 8. x2 – 49y2 = 36
9. 4x2 – y2/9 = 9 10. x2/4 – 9y2 = 100
13. (x – 4)2 – 9(y + 1)2 = 25
17. y2 – 8x – 4x2 + 24y = 29
15. x2 – 6x – 25y2 = 27
18. 9y2 – 18y – 25x2 + 100x = 116
11. x2 – 4y2 + 8y = 5 12. y2 – 8x – 4x2 + 24y = 29
14. 9(x + 2)2 – 4(y + 1)2 = 36
16. x2 – 25y2 – 100y = 200

1.
(0, 3)
(0, -5)
(16, –1)
3.
(3, 0)
(-5, 0)
(-1, -16)
5.
(0, 2)
(0, -6)
(16, -2)
(2, 0)
(-6, 0)
(-2, -16)
7.
(Answers to odd problems) Exercise A.

9.
(-16, -2)
(-16, 6)
(0, 2)
11.
(-2, 0)
(0, -4)
13.
(-16, -2)
(-16, 6)
(0, 2)
15.
(-31, 3)
(-40, 0)

17. 19.
21.
(-4, 64)
(-12, 0) (4, 0)
(-10, 0)
(-6, 2)
(-2, -4)
(0, -8)

1. Center: (0, 0)
x radius: 1
y radius: 1/2
Exercise B.
(-1, 0) (1, 0)
3. Center: (0, 0)
x radius: 1/2
y radius: 3
(0.5, 0)
(-0.5, 0)
Ellipses
5. Center: (0, 0)
x radius: 5
y radius: 3.33
(5, 0)
(-5, 0)
7. Center: (0, 0)
x radius: 10
y radius: 5
(10, 0)
(-10, 0)

9. Center: (0, 0)
x radius: 1.5
y radius: 9
11. Center: (0, 1)
x radius: 1
y radius: 1/2
Ellipses
13. Center: (4, -1)
x radius: 5
y radius: 9/3
15. Center: (3, 0)
x radius:
y radius:
(-1.5, 0) (1.5, 0) (1, 1)
(-1, 1)
(5, 0)
(-5, 0) (9, 0)
(-3, 0)

17. Center: (-1,-12)
x radius: 6.5
y radius: 13
Ellipses
(-1, 1)
(-1, -25)

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