Question

How do you find the foci and sketch the hyperbola `9y^2-x^2=4`?

Answer 1

Please read the explanation.

Explanation:

The standard form for a hyperbola of this type is:

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

where, x and y correspond to any point `(x,y)` on the hyperbola, h and k correspond to the center point, `(h,k)`, "a" is the vertical distance from the center to the vertices, and `sqrt(a^2 + b^2)` is the vertical distance from the center to the foci.

Given: `9y^2 - x^2 = 4`

We will be using algebraic steps to put the above equation into the standard form.

Divide both sides by 4 in the form of `2^2`:

`9y^2/2^2 - x^2/2^2 = 1`

Write the 9 as `3^2` and the denominator of the divisor:

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

Group the denominator of the first term into a single fraction:

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

`sqrt(a^2 + b^2) = sqrt((2/3)^2 + 2^2) = (2sqrt(10))/3`

The foci are located at `(0, -(2sqrt(10))/3) and (0, (2sqrt(10))/3)`

To help you sketch the hyperbola, the center is `(0,0)` and the vertices are at `(0,-2/3) and (0,2/3)`

There is a graph of the original equation: