A hyperbola in the standard (x, y) coordinate plane is defined by the equation: 16x^2 - 9y… — ACT

A hyperbola in the standard $(x, y)$ coordinate plane is defined by the equation:

16x^2 - 9y^2 - 32x + 36y - 164 = 0

What are the coordinates of the foci of this hyperbola?

A
$(1, 7)$ and $(1, -3)$
B
$(6, -2)$ and $(-4, -2)$
$(6, 2)$ and $(-4, 2)$ Answer
D
$(4, -2)$ and $(-6, -2)$
E
$(1 + \sqrt{7}, 2)$ and $(1 - \sqrt{7}, 2)$

Answer

The foci of the hyperbola are

(6, 2)

and

(-4, 2)

The correct answer is the set of coordinates

(6, 2)

and

(-4, 2)

. After rewriting the hyperbola's equation in standard form by completing the square, we obtain

\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1

. The center of the hyperbola is

(1, 2)

. Since the

x^2

term is positive, it has a horizontal transverse axis. The distance from the center to each focus,

c

, satisfies

c^2 = a^2 + b^2 = 9 + 16 = 25

, so

c = 5

. Adding and subtracting this focal distance from the

x

-coordinate of the center yields the foci at

(1 + 5, 2) = (6, 2)

and

(1 - 5, 2) = (-4, 2)

Step-by-Step Solution

Group the

x

-terms and

y

-terms and move the constant to the right side of the equation.

16(x^2 - 2x) - 9(y^2 - 4y) = 164

Grouping the terms allows us to prepare for completing the square for both variables.

Complete the square for

x^2 - 2x

by adding

1

inside the parentheses, and for

y^2 - 4y

by adding

4

inside the parentheses. Add the corresponding balanced quantities to the right side:

16(1) = 16

and

-9(4) = -36

16(x^2 - 2x + 1) - 9(y^2 - 4y + 4) = 164 + 16 - 36

16(x - 1)^2 - 9(y - 2)^2 = 144

This rewrites the quadratic expressions into perfect square binomials.

Divide both sides of the equation by

144

to express it in standard form.

\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1

The standard form of a horizontal hyperbola is

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

, which lets us identify the center,

a

, and

b

directly.

Identify the center

(h, k)

a^2

, and

b^2

, then calculate the focal distance

c

using the relation

c^2 = a^2 + b^2

Center is

(1, 2)

a^2 = 9

and

b^2 = 16

c^2 = 9 + 16 = 25 \implies c = 5

Foci are located at a distance of

c

from the center along the transverse axis.

Determine the coordinates of the foci by shifting the

x

-coordinate of the center by

\pm c

since the transverse axis is horizontal.

Foci coordinates are

(1 \pm 5, 2)

, which gives

(6, 2)

and

(-4, 2)

Adding and subtracting

c

from the center's

x

-coordinate gives the locations of the two foci.

Key Concept

Rewriting a hyperbola equation in standard form by completing the square and finding its foci.

Question

Answer

Step-by-Step Solution

Key Concept

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