Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
511
Note: Problems 22.37–22.40 refer to the hyperbola with equation x
2
– 4y
2
+ 2x + 32y – 27 = 0.
22.39 Write the equations of the asymptotes in slope-intercept form.
One of the asymptotes in Figure 22-10 passes through points (–7,7) and (5,1).
Substitute x
1
= –7, y
1
= 7, x
2
= 5, and y
2
= 1 into the slope formula.
Substitute and x- and y-values from one of the points through which
the asymptote passes (such as x = 5 and y = 1) into the slope-intercept equation
to calculate the y-intercept.
Substitute and into the slope-intercept formula to generate the
equation of the asymptote.
The remaining asymptote has the opposite slope, . Calculate its
y-intercept by substituting its slope and the coordinates of one of the points
through which it passes (such as x = 5 and y = 7) to calculate its y-intercept.
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
512
Substitute and into the slope-intercept formula to generate the
equation of the asymptote.
The equations of the asymptotes are and .
Note: Problems 22.37–22.40 refer to the hyperbola with equation x
2
– 4y
2
+ 2x + 32y – 27 = 0.
22.40 Identify the foci of the hyperbola.
The foci of a hyperbola are the points units away from the center
in both directions along the transverse axis. According to Problem 22.38, a = 3
and b = 6. Calculate c.
The foci of a hyperbola with a vertical transverse axis are (h,k – c) and (h,k + c).
Note: Problems 22.41–22.43 refer to the hyperbola with equation x
2
– 25y
2
– 4x – 21 = 0.
22.41 Write the equation in standard form.
Group the x-terms and move the constant to the right side of the equation.
(x
2
– 4x) – 25y
2
= 0 + 21
Complete the square for the x-expression.
The right side of the standard form equation must be 1, so divide the entire
equation by 25.
The vertices
are above and
below the center of
this hyperbola, so the
foci will be above and
below the center as
well. Unlike ellipses, the
foci of hyperbolas are
farther away from
the center than
the vertices.
Like the foci
of an ellipse with
a vertical major
axis
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
513
Note: Problems 22.41–22.43 refer to the hyperbola with equation x
2
– 25y
2
– 4x – 21 = 0.
22.42 Identify the center and vertices and sketch the graph of the hyperbola.
According to Problem 22.41, the standard form of the hyperbola is
. The positive rational expression left of the equal sign is
written in terms of x, so the transverse axis of the hyperbola is horizontal.
The standard form of a hyperbola with a horizontal transverse axis is
. Therefore, the center of the hyperbola is (h,k) = (2,0),
a = 5, and b = 1.
The vertices of the ellipse are a units away from the center in both directions
along the transverse axis. The transverse axis is horizontal so the vertices are a
units left and right of the center: (h – a,k) and (h + a,k).
Plot the vertices and the endpoints of the conjugate axis, (2,1) and (2,–1).
Draw a rectangle that passes through the endpoints of the axes, as illustrated
by Figure 22-11.
Figure 22-11: The rectangle passes through the endpoints of the transverse axis, (–3,0)
and (7,0), and the endpoints of the conjugate axis, (2,1) and (2,–1).
The asymptotes of the hyperbola extend through opposite corners of the
rectangle in Figure 22-11. The graph passes through the vertices, bends away
from the center point, and approaches (but does not intersect) the asymptotes,
as illustrated by Figure 22-12.
These are the op-
posites of the constants
in the squared quantities.
There’s no constant in the
y
2
expression, so the y-
coordinate of the center
is 0.
a is the
square root of
the positive fraction’s
denominator and b is
the square root of the
negative fraction’s
denominator.
The conjugate
axis is perpendicular
to the transverse axis,
so it’s vertical. The
endpoints are b = 1
unit above and below
the center: (h,k + b)
and (h,k – b).
Chapter Twenty-Two — Conic Sections
The Humongous Book of Algebra Problems
514
Figure 22-12: The graph of x
2
– 25y
2
– 4x – 21 = 0.
Note: Problems 22.41–22.43 refer to the hyperbola with equation x
2
– 25y
2
– 4x – 21 = 0.
22.43 Identify the foci of the hyperbola.
The foci are away from the center in both directions along the
transverse axis. According to Problem 22.42, a = 5 and b = 1. Calculate c.
The transverse axis is horizontal, so the foci are (h – c,k) and (h + c,k).
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