Question

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

Answer 1

`((x--3)^2)/(49) -((y-6)^2)/(32)=1`

Explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

`((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1`

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is
`2a=|4--10|`.

`\implies 2a=|14|`

`\implies 2a=14`

`\implies a=7`

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is
`((-10+4)/(2),(6+6)/(2)=(-3,6)`

We need to use the relation
`a^2+b^2=c^2` to find
`b^2`.

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

`c=|6--3|=9`

`\implies 7^2+b^2=9^2`

`\implies b^2=9^2-7^2`

`\implies b^2=81-49`

`\implies b^2=32`

We substitute these values into the standard equation of the hyperbola to obtain:

`((x--3)^2)/(7^2) - ((y-6)^2)/(32)=1`

`((x+3)^2)/(49) -((y-6)^2)/(32)=1`