Question

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

Answer 1

Please see the explanation.

Explanation:

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

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

The foci are located at the points:

`(h, k - sqrt(a^2 + b^2)) and (h, k + sqrt(a^2 + b^2))`

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

`(y - 0)^2/(1/2)^2 - (x - 0)^2/(1/2)^2 = 1" [2]"`

Substituting into the patterns for the foci:

`(0, 0 - sqrt((1/2)^2 + (1/2)^2)) and (0, 0 + sqrt((1/2)^2 + (1/2)^2))`

The foci are at the points:

`(0, -sqrt(2)/2) and (0, sqrt(2)/2)`

Here is the graph with the foci:

Answer 2

Foci : `(0, +-sqrt2/2)`. See graph and explanation.

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,

`y^2/(1/2)^2-x^2/(1/2)^2=1`, revealing that the hyperbola is

rectangular ( RH ), with [asymptotes]

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

`x^2/(1/4)-y^2/(1/4) =0 rArr y = +-x`

meeting at the center C(0, 0).

The semi axes`a = b = sqrt(1/4) = 1/2`.

The eccentricity e of the RH is `sqrt2`

The Vertices A and A' are `(0, +-1/2)`,

on the ( major ) axis, x = 0.

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

`(0, +-ae) = ( 0, +-sqrt2/2)`

The directrixes DY, DY' are

`y = +-b/e=+-(1/2)/sqrt2=+-1/(2sqrt2)=+-sqrt2/4`.

Now, the RH can be sketched in the order

(i)the guide lines asymptotes `y=+-x`.

(ii) vertices `A(0, 1/2) and A'(-1//2)`

(iii) a few points like `(+-sqrt3/2, +-1)`, near the vertex.