Question

The equation ^2 − 4 − 4^2 + 13 = 0 will produce a hyperbola. How can we tell by simply observing the equation?In what directions do the branches of this hyperbola open? How do you know? Explain. Sketch a graph of this hyperbola, clearly indicating how you have determined thekey characteristics (center, vertices, eccentricity, foci). Give the domain and range of this hyperbola.

Answer 1

we have the equation

`^2−4−4^2+13=0`

Group similar terms and move the constant to the right side

`(^2−4)−4^2=-13`

Complete the square

`\begin{gathered} (y^2-4y+2^2-2^2)-4x^2=-13 \\ (y^2-4y+2^2)-4x^2=-13+2^2 \\ (y^2-4y+2^2)-4x^2=-9 \end{gathered}`

Rewrite as a perfect square

`(y-2)^2-4x^2=-9`

Divide both sides by -9

`\begin{gathered} ((y-2)^2)/(-9)-(4x^2)/(-9)=(-9)/(-9) \\ \\ -((y-2)^2)/(9)+(x^2)/((9)/(4))=1 \\ \\ (x^(2))/((9)/(4))-((y-2)^(2))/(9)=1 \\ \end{gathered}`

The coordinates of the center are (0,2)

The transverse axis is on the x-axis

a^2=9/4 -----------> a=3/2

b^2=9 -----------> b=3

The vertices are --------> (0+1.5,2) and (0-1.5,2)

so

Vertices at (1.5,2) and (-1.5,2)

Find out the value of c

c^2=a^2+b^2

c^2=(9/4)+9

c^2=45/9

c=√5

Find out the coordinates of the foci

(0+√5,2) and (0-√5,2)

using a graphing tool

The domain is the interval (-infinite, -1.5) U (1.5, infinite)

The range is the interval (-infinite, infinite)