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The Humongous Book of Algebra Problems by W. Michael Kelley

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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 ones
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
coefcient (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 youre
actually adding 9(4), so
make sure to add
9(4) = 36 to the
right side of the
equation too.

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