Question

How do you identify the vertices, foci, and direction of `x^2/121-y^2/36=1`?

Answer 1

Given the equation of a hyperbola with a transverse horizontal axis:
`(x-h)^2/a^2-(y-k)^2/b^2=1`
The vertices are: `(h-a,k)` and `(h+a,k)`
The foci are: `(h-sqrt(a^2+b^2),k)` and `(h+sqrt(a^2+b^2),k)`

Explanation:

Note: A horizontal transverse type has a positive x and a negative y term, in a form that is equal to 1, and a vertical transverse type has a negative x term and a positive y term, in a form that is equal to 1.

Given: `x^2/121-y^2/36=1`

Write in the standard form:

`(x-0)^2/11^2-(y-0)^2/6^2=1`

Now, we can write the vertices by observation:

`(-11,0)` and `(11,0)`

Compute `sqrt(a^2+b^2) = sqrt(11^2+6^2)=sqrt(121+36)=sqrt(157)`

Now, we can write the foci with the same ease:

`(-sqrt(157),0)` and `(sqrt(157),0)`