To find: The center, vertices, and foci of the given hyperbola.
It is a vertical hyperbola with center at (−3, −8) , vertices are (−3, −10), (−3, −6) , and the foci are (−3, −8−√53), (−3, −8+√53) .
Given:
(y+8)24−(x+3)249=1 .
Concept used:
The equation of the form (y−k)2a2−(x−h)2b2=1 represents a vertical hyperbola with center at (h, k) , vertices at (h, k±a) , and the foci at (h, k±c) , where c2=a2+b2 .
Calculation:
Consider,
(y+8)24−(x+3)249=1
Since it is in the form of (y−k)2a2−(x−h)2b2=1 , therefore it is a vertical hyperbola.
Here, h=−3, k=−8 .
So, the center is (−3, −8) .
a2=4 , so a=2 .
b2=49 , so b=7 .
Now,
c=√a2+b2=√4+49=√53
So, the foci are
(h, k−c)=(−3, −8−√53) and (h, k+c)=(−3, −8+√53)
And, the vertices are
(h, k−a)=(−3, −8−2) and (h, k+a)=(−3, −8+2)(h, k−a)=(−3, −10) and (h, k+a)=(−3, −6)
Chapter 10 Solutions
High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11
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