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

Dec 22, 2016

#### Explanation:

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

${\left(y - k\right)}^{2} / {a}^{2} - {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [1]}$

The foci are located at the points:

$\left(h , k - \sqrt{{a}^{2} + {b}^{2}}\right) \mathmr{and} \left(h , k + \sqrt{{a}^{2} + {b}^{2}}\right)$

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

${\left(y - 0\right)}^{2} / {\left(\frac{1}{2}\right)}^{2} - {\left(x - 0\right)}^{2} / {\left(\frac{1}{2}\right)}^{2} = 1 \text{ [2]}$

Substituting into the patterns for the foci:

$\left(0 , 0 - \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2}}\right) \mathmr{and} \left(0 , 0 + \sqrt{{\left(\frac{1}{2}\right)}^{2} + {\left(\frac{1}{2}\right)}^{2}}\right)$

The foci are at the points:

$\left(0 , - \frac{\sqrt{2}}{2}\right) \mathmr{and} \left(0 , \frac{\sqrt{2}}{2}\right)$

Here is the graph with the foci:

Dec 22, 2016

Foci : $\left(0 , \pm \frac{\sqrt{2}}{2}\right)$. 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,

${y}^{2} / {\left(\frac{1}{2}\right)}^{2} - {x}^{2} / {\left(\frac{1}{2}\right)}^{2} = 1$, revealing that the hyperbola is

rectangular ( RH ), with [asymptotes]

${x}^{2} / \left(\frac{1}{4}\right) - {y}^{2} / \left(\frac{1}{4}\right) = 0 \Rightarrow y = \pm x$

meeting at the center C(0, 0).

The semi axes$a = b = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

The eccentricity e of the RH is $\sqrt{2}$

The Vertices A and A' are $\left(0 , \pm \frac{1}{2}\right)$,

on the ( major ) axis, x = 0.

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

$\left(0 , \pm a e\right) = \left(0 , \pm \frac{\sqrt{2}}{2}\right)$

The directrixes DY, DY' are

$y = \pm \frac{b}{e} = \pm \frac{\frac{1}{2}}{\sqrt{2}} = \pm \frac{1}{2 \sqrt{2}} = \pm \frac{\sqrt{2}}{4}$.

Now, the RH can be sketched in the order

(i)the guide lines asymptotes $y = \pm x$.

(ii) vertices $A \left(0 , \frac{1}{2}\right) \mathmr{and} A ' \left(- 1 / 2\right)$

(iii) a few points like $\left(\pm \frac{\sqrt{3}}{2} , \pm 1\right)$, near the vertex.