Question

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Answer 1

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

How to graph the hyperbola?

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

Transverse axis

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

The vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

The co-vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola

Read more about hyperbola at:

brainly.com/question/26250569

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