Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
499
Substitute h = 0, k = 3, and r = 5 into the standard form equation.
Ellipses
Major and minor axes, center, foci, and eccentricity
22.22 Identify both standard forms of an ellipse and describe the values of the
constants.
The standard form of an ellipse with a horizontal major axis is
; the standard form of an ellipse with a vertical major
axis is .
The center of the ellipse is (h,k), a represents the distance between the center
and a vertex along the major axis, and b represents the distance between the
center and an endpoint of the minor axis.
Note: Problems 22.23–22.24 refer to the ellipse with center (–1,0), vertical major axis of
length 10, and horizontal minor axis of length 6.
22.23 Graph the ellipse.
If the vertical major axis is ten units long, then it extends five units above and
five units below the center. Similarly, a horizontal minor axis of length six
extends three units left and three units right of the center, as illustrated by
Figure 22-5.
Figure 22-5: This ellipse has the center (–1,0), a major axis ten units long, and a
minor axis six units long.
The axes
of an ellipse are
the perpendicular
segments that pass
through the center
and extend to the
right, left, top, and
bottom “edges” of the
ellipse. The longer of
the two is called the
major axis, and
the other one’s
the minor axis.
Five units
up = (–1,5); ve units
down = (–1,–5); three
units left = (–4,0);
and three units
right = (2,0).
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
500
Note: Problems 22.23–22.24 refer to the ellipse with center (–1,0), vertical major axis of
length 10, and horizontal minor axis of length 6.
22.24 Write the equation of the ellipse in standard form.
According to Problem 22.22, the standard form of an ellipse with a vertical
major axis is . The center of the ellipse is (–1,0), so
h = –1 and k = 0. The values of a and b are half of the lengths of the major and
minor axes, respectively. Therefore, a = 10 ÷ 2 = 5 and b = 6 ÷ 2 = 3. Substitute h,
k, a, and b into the standard form equation.
Note: Problems 22.25–22.27 refer to the ellipse with equation x
2
+ 9y
2
– 6x – 36y + 36 = 0.
22.25 Write the equation in standard form.
Group the x-terms together in a set of parentheses, group the y-terms in a
second set of parentheses, and move the constant to the right side of the
equation.
(x
2
– 6x) + (9y
2
– 36y) = 0 – 36
In order to complete the square for the expression 9y
2
– 36y, the coefficient of
the squared term must be 1.
(x
2
– 6x) + 9(y
2
– 4y) = –36
Complete the square for each of the parenthetical quantities.
The equation of an ellipse in standard form contains only the constant 1 on
the right side of the equation. Divide the entire equation by 9, the constant
currently right of the equal sign.
a
2
appears
beneath the y-
expression when the
major axis is vertical
and appears beneath
the x-expression when
the major axis is
horizontal.
But it’s not,
so factor the
coefcient (9) out of
both y-terms.
To complete
the square, you
have to add 9 to
the x-parentheses
and 4 to the y-
parentheses. However,
the y-parentheses are
multiplied by 9, so you’re
actually adding 9(4), so
make sure to add
9(4) = 36 to the
right side of the
equation too.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
501
Note: Problems 22.25–22.27 refer to the ellipse with equation x
2
+ 9y
2
– 6x – 36y + 36 = 0.
22.26 Graph the ellipse.
According to Problem 22.25, the standard form of the ellipse is
. The center of the ellipse is (3,2). The constant beneath
the x-expression (9) is greater than the constant beneath the y-expression (1),
so the major axis of the ellipse is horizontal. Let a
2
be the greater of the two
denominators and b
2
be the lesser of the denominators: a
2
= 9 and b
2
= 1, so a = 3
and b = 1.
To graph the ellipse, plot the points three units right and left of the center
(because a = 3 and the major axis is horizontal) and the points one unit above
and below the center (because b = 1 and the minor axis is vertical). The ellipse
is graphed in Figure 22-6.
Figure 22-6: The graph of the ellipse x
2
+ 9y
2
– 6x – 36y + 36 = 0 has center (3,2), a
horizontal major axis of length 6, and a vertical minor axis of length 2.
The x-
coordinate
of the center
is the opposite of
the number in the x-
expression (x – 3)
2
. The
y-coordinate of the
center is the opposite
of the constant in
the y-expression
(y – 2)
2
.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
502
Note: Problems 22.25–22.27 refer to the ellipse with equation x
2
+ 9y
2
– 6x – 36y + 36 = 0.
22.27 Identify the vertices of the ellipse.
The vertices of an ellipse are the endpoints of the major axis. Consider the
graph of the ellipse in Figure 22-6. The major axis is the horizontal segment
with endpoints (0,2) and (6,2), so those points are the vertices of the ellipse.
Note: Problems 22.28–22.30 refer to the ellipse with equation 9x
2
+ 4y
2
+ 90x – 48y + 333 = 0.
22.28 Write the equation in standard form.
Group the x-terms as one quantity, the y-terms as another quantity, and move
the constant to the right side of the equation.
(9x
2
+ 90x) + (4y
2
– 48y) = 0 – 333
Factor the coefficients of the squared terms out of each quantity.
9(x
2
+ 10x) + 4(y
2
– 12y) = –333
Complete the square for each quantity.
Divide the entire equation by 36 to set the right side of the equation equal to 1.
Note: Problems 22.28–22.30 refer to the ellipse with equation 9x
2
+ 4y
2
+ 90x – 48y + 333 = 0.
22.29 Identify the center, vertices, and foci of the ellipse.
The denominator beneath the y-expression is greater than the denominator
beneath the x-expression, so the major axis of the ellipse is vertical, a
2
= 9,
and b
2
= 4. Thus a = 3 and b = 2. The center of the ellipse is comprised of the
opposites of the constants within the squared expressions: (h,k) = (–5,6).
Because the major axis is vertical, the vertices are the points a = 3 units above
and below the center.
If the major
axis is horizontal, the
vertices are (h – a,k)
and (h + a,k).
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
503
The foci are the two points c units away from the center along the major axis
such that . Calculate c.
Calculate the coordinates of the foci.
Note: Problems 22.28–22.30 refer to the ellipse with equation 9x
2
+ 4y
2
+ 90x – 48y + 333 = 0.
22.30 Calculate the eccentricity of the ellipse.
The eccentricity of an ellipse is defined as , where c is the distance
between the center and a focus and a is the distance between the center and a
vertex. According to Problem 22.29, a = 3 and .
Note: Problems 22.31–22.32 refer to the ellipse with center (4,–1) that passes through the
origin if the major axis is horizontal and ab = 8.
22.31 Write the equation of the ellipse in terms of a.
The standard form of an ellipse with a horizontal major axis is
. Solve the equation ab = 8 for b.
Substitute h = 4, k = –1, and into the equation and simplify.
If you go
up and down
from the center to
reach the vertices,
you also go up and
down (but not quite
as far) to reach
the foci.
If the
major axis
is horizontal, the
foci are (h – c,k)
and (h + c,k).
Simplify the
complex fraction by
multiplying the top and
bottom by the reciprocal
of the denominator:
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