Holt Mcdougal Larson Algebra 2: Student Edition 2012
1st Edition
ISBN: 9780547647159
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Expert Solution & Answer
Chapter 8.5, Problem 25E
Solution
Write the standard form of the equation of the hyperbola with the given characteristics.
x24−y250=1
Given information:
Foci at (−3√6,0),(3√6,0) ,
Vertices at (−2,0),(2,0) .
Formula used:
- Standard form of equation of a hyperbola with foci (±c,0) and vertices (±a,0) on the x -axis is given by:
- Standard form of equation of a hyperbola with foci (0,±c) and vertices (0,±a) on the y -axis is given by:
- Focus of hyperbola with vertices as (±a,0) or (0,±a) and co- vertices (0,±b) or (±b,0) is found as:
x2a2−y2b2=1
y2a2−x2b2=1
c2=a2+b2
Calculation:
In order to write the standard form of the equation of the hyperbola with given foci (−3√6,0),(3√6,0) and vertices (−2,0),(2,0) first note that both foci and vertices are on x -axis equidistance from the center, so the transverse axis is horizontal and center is the origin.
Thus, the equation of the hyperbola will be of the form
x2a2−y2b2=1 , with vertices as (±a,0) and co- vertices (0,±b) .
Since, foci at 3√6 units i.e. c=3√6 and vertices at 2 units each i.e. a=2 , so calculate the co-vertices, using the formula
c2=a2+b2(3√6)2=(2)2+b29×6=4+b2b2=54−4b2=50b=√50b=5√2
Putting, a=2 and b=5√2 in the above form of hyperbola:
x222−y2(5√2)2=1x24−y225×2=1x24−y250=1
Thus, the standard form of the equation of hyperbola with given information is:
x24−y250=1
Chapter 8 Solutions
Holt Mcdougal Larson Algebra 2: Student Edition 2012
Ch. 8.1 - Prob. 1GPCh. 8.1 - Prob. 2GPCh. 8.1 - Prob. 3GPCh. 8.1 - Prob. 4GPCh. 8.1 - Prob. 5GPCh. 8.1 - Prob. 6GPCh. 8.1 - Prob. 1ECh. 8.1 - Prob. 2ECh. 8.1 - Prob. 3ECh. 8.1 - Prob. 4E
Ch. 8.1 - Prob. 5ECh. 8.1 - Prob. 6ECh. 8.1 - Prob. 7ECh. 8.1 - Prob. 8ECh. 8.1 - Prob. 9ECh. 8.1 - Prob. 10ECh. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.1 - Prob. 33ECh. 8.1 - Prob. 34ECh. 8.1 - Prob. 35ECh. 8.1 - Prob. 36ECh. 8.1 - Prob. 37ECh. 8.1 - Prob. 38ECh. 8.1 - Prob. 39ECh. 8.1 - Prob. 40ECh. 8.1 - Prob. 41ECh. 8.1 - Prob. 42ECh. 8.1 - Prob. 43ECh. 8.1 - Prob. 44ECh. 8.1 - Prob. 45ECh. 8.1 - Prob. 46ECh. 8.1 - Prob. 47PSCh. 8.1 - Prob. 48PSCh. 8.1 - Prob. 49PSCh. 8.1 - Prob. 50PSCh. 8.1 - Prob. 51PSCh. 8.1 - Prob. 52PSCh. 8.1 - Prob. 53PSCh. 8.1 - Prob. 54PSCh. 8.1 - Prob. 55PSCh. 8.1 - Prob. 56PSCh. 8.1 - Prob. 57PSCh. 8.2 - Prob. 1GPCh. 8.2 - Prob. 2GPCh. 8.2 - Prob. 3GPCh. 8.2 - Prob. 4GPCh. 8.2 - Prob. 5GPCh. 8.2 - Prob. 6GPCh. 8.2 - Prob. 7GPCh. 8.2 - Prob. 8GPCh. 8.2 - Prob. 9GPCh. 8.2 - Prob. 1ECh. 8.2 - Prob. 2ECh. 8.2 - Prob. 3ECh. 8.2 - Prob. 4ECh. 8.2 - Prob. 5ECh. 8.2 - Prob. 6ECh. 8.2 - Prob. 7ECh. 8.2 - Prob. 8ECh. 8.2 - Prob. 9ECh. 8.2 - Prob. 10ECh. 8.2 - Prob. 11ECh. 8.2 - Prob. 12ECh. 8.2 - Prob. 13ECh. 8.2 - Prob. 14ECh. 8.2 - Prob. 15ECh. 8.2 - Prob. 16ECh. 8.2 - Prob. 17ECh. 8.2 - Prob. 18ECh. 8.2 - Prob. 19ECh. 8.2 - Prob. 20ECh. 8.2 - Prob. 21ECh. 8.2 - Prob. 22ECh. 8.2 - Prob. 23ECh. 8.2 - Prob. 24ECh. 8.2 - Prob. 25ECh. 8.2 - Prob. 26ECh. 8.2 - Prob. 27ECh. 8.2 - Prob. 28ECh. 8.2 - Prob. 29ECh. 8.2 - Prob. 30ECh. 8.2 - Prob. 31ECh. 8.2 - Prob. 32ECh. 8.2 - Prob. 33ECh. 8.2 - Prob. 34ECh. 8.2 - Prob. 35ECh. 8.2 - Prob. 36ECh. 8.2 - Prob. 37ECh. 8.2 - Prob. 38ECh. 8.2 - Prob. 39ECh. 8.2 - Prob. 40ECh. 8.2 - Prob. 41ECh. 8.2 - Prob. 42ECh. 8.2 - Prob. 43ECh. 8.2 - Prob. 44ECh. 8.2 - 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