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

2 Answers
Dec 22, 2016

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:
Desmos.com

Dec 22, 2016

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.