x^2-9y^2+36y-72 = 0

Question

find the center , vertices,foci,and asymptotes of the hyperbola, and sketch the graph.

Answer 1

Given equation x^2-9y^2+36y-72 = 0

x^2-9y^2+36y-72 = 0

x^2-9y^2+36y-36-36 = 0

x^2-9(y^2-4y+4)-36 = 0

x^2-9(y-2)^2-36= 0

(x-0)^2-9(y-2)^2 = 36

Divide to each side by 36.

(x-0)^2/36-9(y-2)^2/36 = 36/36

(x-0)^2/36-(y-2)^2/4 = 1

(x-0)^2/6^2-(y-2)^2/2^2 = 1

a = 6, b = 2, h = 0, k = 2

→ Center = (h,k) = (0,2)

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

(0+6,2) = (6,2)

(0-6,2) = (-6,2)

→ Vertices (6,2), (-6,2)

c = √(a^2+b^2) = √(36+4) = √(40)

 Foci = (h+c,k),(h-c,k)

(0+√40,2), (0-√40,2)

(√40,2), (-√40,2)

→ Foci (6.32,2), (-6.32,2)

Answer 2

The hyperbola equation is

To change the expressions  (y ²  - 4y) into a perfect square trinomial, add (half the y coefficient)² to each side of the equation.

The standard form for hyperbola is in a form = 1, So divide both sides of equation by 36 to set it equal to 1.

Compare it to standard form of hyperbola

"a " is the number in the denominator of the positive term

If the x -term is positive, then the hyperbola is horizontal

a = semi-transverse axis = 6

b = semi-conjugate axis = 2

center: (h, k ) = (0,2)

Vertices: (h + a, k ), (h - a, k )

= (0+6, 2) , (0-6, 2)

Vertices of the hyperbola  (6, 2), (-6, 2)

c  = distance from the center to each focus along the transverse axis

Foci: (h + c, k ), (h - c, k )

Foci of the hyperbola

The equations for asymptotes are 

Asymptotes of hyperbola are .

Answer 3

Continuous...

Graph

• Step-1

Plot the center point of hyperbola.

• Step-2

The a  value is the horizontal distancse from the center to the edge of box.

The b  value is the vertical distance from thecenter to the edge of the box.

In this case a  = 6, b  = 2 Draw in the the box these values.

• Step-3

Asymptotes of hyperbola are

Draw slant asympototes (remember that always pass through the corners of box)

• Step-4

Complete the hyperbola (The vertices will touch the box)