SOLUTION: Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11). A.y squared over forty minus x squared over eighty one = 1 B.y square

Algebra ->  Trigonometry-basics -> SOLUTION: Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11). A.y squared over forty minus x squared over eighty one = 1 B.y square      Log On


   

Question 1136096: Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11).

A.y squared over forty minus x squared over eighty DISABLED_event_one= 1

B.y squared over eighty one minus x squared over one hundred and twenty DISABLED_event_one= 1

C.y squared over eighty one minus x squared over forty = 1

D.y squared over one hundred and twenty one minus x squared over eighty DISABLED_event_one= 1
Answer by MathLover1(20850) About Me  (Show Source):
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Find an equation in standard form for the hyperbola with
vertices at (0, ±9) and
foci at (0, ±11).
The standard form of the equation for this type of hyperbola given the center is at the origin is:

y%5E2+%2F+a%5E2+-+x%5E2+%2F+b%5E2+=1+
Your coordinates of the vertices are (0, ±9) and (0, ±11) making a=9 and c=11

c%5E2=a%5E2%2Bb%5E2
11%5E2=9%5E2%2Bb%5E2
b%5E2=121-81
b%5E2=40
so your equation is:
y%5E2+%2F+81+-+x%5E2+%2F+40+=1+

answer: C.y squared over eighty one minus x squared over forty = 1