FROM THE GIVEN INFORMATION FIND EQUATION OF HYPERBOLA

Find the standard form of the equation of each hyperbola.

Problem 1 :

Foci (0, ±4), vertices (0, ±2)

Solution:

 Foci = (0, ±4)

= (0, ±c)

c = 4

vertices = (0, ±a)

= (0, ±2)

a = 2

b2 = c2 - a2

= 42 - 22

= 16 - 4

b2 = 12

So, the equation of hyperbola is

y2a2-x2b2=1y24-x212=1

Problem 2 :

Vertices (±4, 0), Asymptotes: y = ±3x

Solution:

Vertices (4, 0) (-4, 0)

The hyperbola is the horizontal transverse axis type.

x2a2-y2b2=1

Vertices = (±a, 0)

a = 4

slope of asymptotes=±ba±b4=±3b=12

So, the equation of hyperbola is

x242-y2122=1x216-y2144=1

Problem 3 :

Endpoints of transverse axis: (±6, 0), Asymptotes: y = ±2x

Solution:

The hyperbola is the horizontal transverse axis type.

x2a2-y2b2=1

Here a = 6

slope of asymptotes=±ba±b6=±2b=12

So, the equation of hyperbola is

x262-y2122=1x236-y2144=1

Problem 4 :

Foci (0, ±3), length of transverse axis 2

Solution:

Foci = (0, ±c)

c = 3

length of transverse axis 2a = 2

a = 1

b2 = c2 - a2

= 32 - 12

= 9 - 1

b2 = 8

So, the equation of hyperbola is

y2a2-x2b2=1y21-x28=1

Problem 5 :

find-hyperboal-q5

Solution:

From given graph, 

Center (h, k) = (2, 1)

Vertices = (4, 1) (0, 1)

a = 2

b = 3

So, the equation of hyperbola is

(x-h)2a2-(y-k)2b2=1(x-2)222-(y-1)232=1(x-2)24-(y-1)29=1

Problem 6 :

find-hyperboal-q6.png

Solution:

From given graph, 

Center (h, k) = (0, 0)

Vertices = (3, 0) (-3, 0)

a = 3

b = 1

So, the equation of hyperbola is

(y-k)2a2-(x-h)2b2=1(y-0)232-(x-0)212=1y29-x21=1

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