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.

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