PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
Chapter 8.3, Problem 36E
Solution
An equation in standard form for the hyperbola that satisfies the given conditions.
It has been determined that the equation of the given hyperbola in standard form is 4(y+112)249−(x+3)218=1 .
Given:
Transverse axis end points (−3,−9) and (−3,−2) ; foci (−3,−11) and (−3,0) .
Concept used:
The equation of a hyperbola in standard form with focal axis y=k , foci at (h±c,k) , vertices at (h±a,k) , semi-transverse axis of length a , semi-conjugate axis of length b and asymptotes y=±ba(x−h)+k ; is (x−h)2a2−(y−k)2b2=1 , where c=√a2+b2 .
The equation of a hyperbola in standard form with focal axis x=h , foci at (h,k±c) , vertices at (h,k±a) , semi-transverse axis of length a , semi-conjugate axis of length b and asymptotes y=±ab(x−h)+k ; is (y−k)2a2−(x−h)2b2=1 , where c=√a2+b2 .
Calculation:
It is given that the transverse axis end points of the given hyperbola are (−3,−9) and (−3,−2) .
That is, the vertices of the given hyperbola are (−3,−9) and (−3,−2) .
Comparing these vertices with (h,k±a) , it follows that:
h=−3 , k+a=−2 and k−a=−9 .
Adding the equations; k+a=−2 and k−a=−9 , it follows that,
2k=−11
Simplifying,
k=−112
Put k=−112 in k+a=−2 to get a=72 .
It is also given that foci of the given hyperbola are (−3,−11) and (−3,0) .
Comparing these foci with (h,k±c) , it follows that:
h=−3 , k+c=0 and k−c=−11 .
Put k=−112 in k+c=0 to get c=112 .
Now, put c=112 and a=72 in c=√a2+b2 to get,
112=√(72)2+b2(112)2=(72)2+b2b2=(112)2−(72)2b=±√(112)2−(72)2
Simplifying,
b=±√(112)2−(72)2=±12√(11)2−(7)2=±12√(11−7)(11+7)=±12√(4)(18)
Solving,
b=±12(2)(3)√2=±3√2
So, b=±3√2 .
Put h=−3 , k=−112 , a=72 and b=±3√2 in (y−k)2a2−(x−h)2b2=1 to get,
(y−(−112))2(72)2−(x−(−3))2(±3√2)2=1
Simplifying,
4(y+112)249−(x+3)218=1
This is the equation of the given hyperbola in standard form.
Conclusion:
It has been determined that the equation of the given hyperbola in standard form is 4(y+112)249−(x+3)218=1 .
Chapter 8 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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