Question

Write the equation for the hyperbola with foci (1, -5) (9,-5), and conjugate axis of length 6

Answer 1

Answer:

Answer

The answer would be A. 1 1/5

48/40 = 1 8/40

1 8/40 Simplify to 1 1/5

Hope this helps!

Step-by-step explanation:

Answer 2

The required equation of hyperbola is $(y+5)^2/20 + (x-5)^2 /16 = 1$,

Foci and conjugate axis length are given as (1, -5) (9,-5) and 6 respectively.

What is hyperbola?

Hyperbola is cross section cut out from the cone.

Here, the standard equation of hyperbola is $= (x-h)^2/a^2+(y-k)^2/b^2=1$

Point (h ,k) = (1+9/2, -5-5/2)
                 = (5,-5)
Now,
2a = 8
a² = 16
and also c² = 36 (conjugate axis)
c²=a²+b²
36 = 16 + b²
b² = 20

Put this values in standard equation.
$(y+5)^2/20 + (x-5)^2 /16 = 1$

Thus, The required equation of hyperbola is $(y+5)^2/20 + (x-5)^2 /16 = 1$.

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