Question

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

Answer 1

See explanation and graph.

Explanation:

TIn the standard form the equation is

`y^2/(1/2)^2-x^2/1^1=1`

Center C is the origin (0, 0)

Major axis : y-axis, x = 0.

Transverse axis : x-axis, y = 0.

Semi major axis `a = 1/2`

Semi transverse axis `b = 1`

Eccentricity : `e =sqrt(1+a^2/b^2)=sqrt5/2`

Vertices A and A' on major axis x = 0 : `(0, +-a) = (0, +-1/2)`-

Foci S and S' on major axis : (`0, +-ae)=(0, +-sqrt5)`

The asymptotes : `4y^2-x^2=0`, giving `y=+-x/sqrt2`.

The Socratic graph is inserted.

graph{(4y^2-x^2-1)(4y^2-x^2)((y-sqrt5)^2+x^2-.01)((y+sqrt5)^2+x^2-.01)((y-.5)^2+x^2-.01)((y+.5)^2+x^2-.01)=0 [-5, 5, -2.5, 2.5]}