
Finding the vertices, foci and asymptotes of a hyperbola
Interactive Video
•
Mathematics
•
11th Grade - University
•
Practice Problem
•
Hard
Wayground Content
FREE Resource
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7 questions
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1.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the center of the hyperbola given by the equation X - 1^2 / 4 - Y + 2^2 / 1 = 1?
(1, 2)
(-1, -2)
(-1, 2)
(1, -2)
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
How do you determine if a hyperbola has a horizontal or vertical transverse axis?
By the coefficient of X
By the sign of the equation
By the value of B^2
By which variable is above A^2
3.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What are the coordinates of the vertices for the hyperbola with center (1, -2) and a^2 = 4?
(3, 2) and (-1, 2)
(0, -2) and (2, -2)
(1, 0) and (1, -4)
(3, -2) and (-1, -2)
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Which formula is used to calculate the distance to the foci of a hyperbola?
C^2 = a^2 - b^2
C = a - b
C^2 = a^2 + b^2
C = a + b
5.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the approximate value of sqrt(5) used to find the foci?
2.2
2.5
2.8
2.0
6.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What is the equation for the asymptotes of a hyperbola with a horizontal transverse axis?
y = h ± (a/b)(x - k)
y = h ± (b/a)(x - k)
y = k ± (a/b)(x - h)
y = k ± (b/a)(x - h)
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
What are the equations of the asymptotes derived in the video?
y = 1/2x + 3/2 and y = -1/2x + 5/2
y = 1/2x - 3/2 and y = -1/2x - 5/2
y = 1/2x + 5/2 and y = -1/2x + 3/2
y = 1/2x - 5/2 and y = -1/2x - 3/2
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