SOLUTION: Which hyperbolas have one vertex in common with the hyperbola (y-4)^2/7^2-(x+6)^2/8^2=1? a. (y-8)^2/5^2-(x+6)^2/6^2=1 b. (x-1)^2/8^2-(y-11)^2/7^2=1 c. (x+15)^2/9^2-(y+3)^2/5

Algebra ->  Finance -> SOLUTION: Which hyperbolas have one vertex in common with the hyperbola (y-4)^2/7^2-(x+6)^2/8^2=1? a. (y-8)^2/5^2-(x+6)^2/6^2=1 b. (x-1)^2/8^2-(y-11)^2/7^2=1 c. (x+15)^2/9^2-(y+3)^2/5      Log On


   

Question 1180216: Which hyperbolas have one vertex in common with the hyperbola
(y-4)^2/7^2-(x+6)^2/8^2=1?
a. (y-8)^2/5^2-(x+6)^2/6^2=1
b. (x-1)^2/8^2-(y-11)^2/7^2=1
c. (x+15)^2/9^2-(y+3)^2/5^2=1
d. (y-6)^2/6^2-(x+6)^2/4^2=1
e. (y-23)^2/12^2-(x+6)^2/8^2=1
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Which hyperbolas have one vertex in common with the hyperbola
%28y-4%29%5E2%2F7%5E2-%28x%2B6%29%5E2%2F8%5E2=1
Up-down hyperbola with center (h, k )= (-6, 4), a=7, b=8
The vertices:
(h, k%2Ba ), (h, k-a )
=>(-6, 4%2B7 ), (-6, 4-7 )
=>(-6,+11 ), (-6, -3 )


a.
+%28y-8%29%5E2%2F5%5E2-%28x%2B6%29%5E2%2F6%5E2=1
center (h, k )= (-6, 8), a=5, b=5
The vertices:
(h, k%2Ba ), (h, k-a )
=>(-6, 8%2B5 ), (-6, 8-5 )
=>(-6,+13 ), (-6, 3 )

b.

%28x-1%29%5E2%2F8%5E2-%28y-11%29%5E2%2F7%5E2=1
Right-left hyperbola with (h,+k )= (1, 11+), a=8, b=7
The vertices
(h%2Ba, k+), (h-a,+k )
=>(1%2B8, 11+), (1-8,+11 )
=>(9, 11+), (-7,+11 )

c.
%28x%2B15%29%5E2%2F9%5E2-%28y%2B3%29%5E2%2F5%5E2=1
Right-left hyperbola with (h,+k )= (-15, -3+), a=9, b=5
The vertices
(h%2Ba, k+), (h-a,+k )
=>(-15%2B9, -3+), (-15-9,+-3 )
=>(-6, -3+), (-24,+-3 )

d.

%28y-6%29%5E2%2F6%5E2-%28x%2B6%29%5E2%2F4%5E2=1
Up-down hyperbola with center (h, k )= (-6, 6), a=6, b=4
The vertices:
(h, k%2Ba ), (h, k-a )
=>(-6, 6%2B6 ), (-6, 6-6 )
=>(-6,+12 ), (-6, 0 )

e.

%28y-23%29%5E2%2F12%5E2-%28x%2B6%29%5E2%2F8%5E2=1
Up-down hyperbola with center (h, k )= (-6, 23), a=12, b=8
The vertices:
(h, k%2Ba ), (h, k-a )
=>(-6, 23%2B12 ), (-6, 23-12 )
=>(-6,+35 ), (-6, 11 )

hyperbolas which have one vertex in common with the given hyperbola:
c has =>(-6, -3+)
e has =>(-6, 11 )