# How do you find the foci and sketch the hyperbola #4y^2-4x^2=1#?

##### 2 Answers

Please see the explanation.

#### Explanation:

The standard form for the equation of a hyperbola with a vertical transverse axis is:

The foci are located at the points:

Let's write your equation in the form of equation [1]:

Substituting into the patterns for the foci:

The foci are at the points:

Here is the graph with the foci:

Foci :

#### Explanation:

graph{(4y^2-4x^2-1)(x^2-y^2)(x^2+(y-0.707)^2-.001)(x^2+(y+0.707)^2-.001)=0 [-2.5, 2.5, -1.25, 1.25]}

In the standard form,

rectangular ( RH ), with [asymptotes]

(https://socratic.org/precalculus/functions-defined-and-notation/asymptotes)

meeting at the center C(0, 0).

The semi axes

The eccentricity e of the RH is

The Vertices A and A' are

on the ( major ) axis, x = 0.

The foci S and S' on the major y- axis are

The directrixes DY, DY' are

Now, the RH can be sketched in the order

(i)the guide lines asymptotes

(ii) vertices

(iii) a few points like