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Algebra II : Hyperbolic Functions

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Example Question #11 : Hyperbolic Functions

Which of the following shapes does the graph of the equation [(x-6)^2]/36 + [(y+2)^2]/25 = 1 take?

Possible Answers:

Hyperbola

Circle

Parabola

Ellipse

Correct answer:

Ellipse

Explanation:

The equation for a horizontal hyperbola is [(x-h)^2]/(a^2) - [(y-v)^2]/(b^2) = 1 . The equation for a vertical hyperbola is [(y-v)^2]/(b^2) - [(x-h)^2]/(a^2) = 1 . Both include x- and y-terms combined using subtraction. The equation for an ellipse is [(x-h)^2]/(a^2) + [(y-k)^2]/(b^2) = 1 . Because the given equation connects the x- and y-terms using addition rather than subtraction, it represents an ellipse rather than a hyperbola. If the equation took the form (x - h)^2 + (y - k)^2 = r^2 , it would represent a circle. If the equation took the form y = a(x-h)^2 + k , it would represent a parabola.

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Example Question #11 : Hyperbolic Functions

Which of the following shapes does the graph of the equation [(x-3)^2]/100 - [(y-4)^2]/144 = 1 take?

Possible Answers:

Circle

Hyperbola

Ellipse

Parabola

Correct answer:

Hyperbola

Explanation:

The equation for a horizontal hyperbola is [(x-h)^2]/(a^2) - [(y-v)^2]/(b^2) = 1 . The equation for a vertical hyperbola is [(y-v)^2]/(b^2) - [(x-h)^2]/(a^2) = 1 . Both include x- and y-terms combined using subtraction. If the equation took the form [(x-h)^2]/(a^2) + [(y-k)^2]/(b^2) = 1 (using addition rather than subtraction to combine the x- and y-terms), it would represent an ellipse. If the equation took the form (x - h)^2 + (y - k)^2 = r^2 , it would represent a circle. If the equation took the form y = a(x-h)^2 + k , it would represent a parabola.

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Example Question #13 : Hyperbolic Functions

What are the coordinates of the center of the hyperbolic inequality $\frac{[(y+2)^{2}]}{9}-\frac{[(x-1)^{2}]}{4}\geq 1$ ?

Possible Answers:

(-2, 1)

(2, -1)

(-1, 2)

(1, -2)

Correct answer:

(1, -2)

Explanation:

The equation for a horizontal hyperbola is $\frac{[(x-h)^{2}]}{(a^{2})}-\frac{[(y-v)^{2}]}{(b^{2})}=1$ . The equation for a vertical hyperbola is $\frac{[(y-v)^{2}]}{(b^{2})}-\frac{[(x-h)^{2}]}{(a^{2})}=1$ . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. Because the two terms are combined using subtraction and the y-term appears first, this inequality represents a vertical hyperbola. To derive the center of a hyperbola from its equation or inequality, flip the sign of the constants that appear after the x and y in the equation or inequality. The constant following x is -1, so the x-coordinate of the center is 1. The constant following y is 2, so the y-coordinate of the center is -2.

The graph of the hyperbolic inequality $\frac{[(y+2)^{2}]}{9}-\frac{[(x-1)^{2}]}{4}\geq 1$ appears as follows:

Hyperbola2

With vertices of (1, 1) and (1, -5), you can see that the midpoint between them is (1, -2).

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Example Question #14 : Hyperbolic Functions

What are the coordinates of the center of the hyperbolic inequality $\frac{[(x+3)^{2}]}{9}-\frac{(y^{2})}{4}< 1$ ?

Possible Answers:

(-3, 0)

(4, -3)

(0, 3)

(3, 4)

Correct answer:

(-3, 0)

Explanation:

The equation for a horizontal hyperbola is $\frac{[(x-h)^{2}]}{(a^{2})}-\frac{[(y-v)^{2}]}{(b^{2})}=1$ . The equation for a vertical hyperbola is $\frac{[(y-v)^{2}]}{(b^{2})}-\frac{[(x-h)^{2}]}{(a^{2})}=1$ . In both, (h, v) is the center of the hyperbola. Hyperbolic inequalities use an inequality sign rather than an equals sign, but otherwise have the same form as hyperbolic equations. Because the two terms are combined using subtraction and the x-term appears first, this inequality represents a horizontal hyperbola. To derive the center of a hyperbola from its equation or inequality, flip the sign of the constants that appear after the x and y in the equation or inequality. The constant following x is 3, so the x-coordinate of the center is -3. No constant follows y, so the y-coordinate of the center is 0.

The graph of the hyperbolic inequality $\frac{[(x-h)^{2}]}{(a^{2})}-\frac{[(y-v)^{2}]}{(b^{2})}=1$ appears as follows:

Hyperbola1

As you can see, the midpoint between the two vertices of (-6, 0) and (0, 0) is (-3, 0).

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Example Question #15 : Hyperbolic Functions

Which equation does this graph represent?

Screen shot 2020 08 26 at 8.10.13 am

Possible Answers:

[(x+2)^2]/1 - [(y+1)^2]/4 = 1

[(x+2)^2]/1 + [(y+1)^2]/4 = 1

[(y+1)^2]/4 - [(x+2)^2]/1 = 1

[(y+1)^2]/4 + [(x+2)^2]/1 = 1

Correct answer:

[(x+2)^2]/1 - [(y+1)^2]/4 = 1

Explanation:

The equation for a horizontal hyperbola is [(x-h)^2]/(a^2) - [(y-v)^2]/(b^2) = 1 . The equation for a vertical hyperbola is [(y-v)^2]/(b^2) - [(x-h)^2]/(a^2) = 1 . In both, (h, v) is the center of the hyperbola. The graph shows a horizontal hyperbola, so in its corresponding equation the x-term must appear first. The center lies at (-2, -1), so x must be followed by the constant 2, and y must be followed by the constant 1. The graph shows a hyperbola rather than an ellipse, so the x- and y-terms must be combined using subtraction rather than addition.

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Example Question #12 : Hyperbolic Functions

In which direction does the graph of the hyperbola [(x-3)^2]/100 - [(y-4)^2]/144 = 1 open?

Possible Answers:

vertical

horizontal

Correct answer:

horizontal

Explanation:

The equation for a horizontal hyperbola is [(x-h)^2]/(a^2) - [(y-v)^2]/(b^2) = 1 . The equation for a vertical hyperbola is [(y-v)^2]/(b^2) - [(x-h)^2]/(a^2) = 1 . The x-term appears first, so the given equation represents a horizontal hyperbola.

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Algebra II : Hyperbolic Functions

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #11 : Hyperbolic Functions

Example Question #11 : Hyperbolic Functions

Example Question #13 : Hyperbolic Functions

Example Question #14 : Hyperbolic Functions

Example Question #15 : Hyperbolic Functions

Example Question #12 : Hyperbolic Functions

All Algebra II Resources

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