# How do you graph & find the focus as well as the vertex of #4y^2 - 9x^2 - 40y - 54x - 17 = 0#?

##### 1 Answer

Use the standard form for a hyperbola of this type:

The vertices are:

And the foci are:

#### Explanation:

Given:

Here is a graph of equation [1].

graph{4y^2 - 9x^2 - 40y - 54x - 17 = 0 [-20, 20, -6, 16]}

The standard form for a hyperbola of this type is:

We want to make equation [1] look like equation [2], because we know that:

The vertices are:

And the foci are:

Group the x terms and y terms together and move the constant to the right:

We want to complete the squares in equation [1.1], therefore, we shall add

Remove a common factor of 4 from the y terms and a common factor of -9 from the x terms:

From the expansion

Into equation [1.3], substitute

From the expansion

Into equation [1.4], substitute

Simplify the right side of equation [1.5]:

Divide both sides of the equation by 36:

Write the denominators as squares:

Using the pattern for the vertices,

The vertices are:

Using the pattern for the foci,

The foci are: