Question

Which is the equation of a hyperbola centered at the origin with focus 0,4) and vertex (0, square root of 12 )?

Answer 1

The equation of the hyperbola is:

`(x^(2))/(76) - (y^(2))/(12) = 1`

Explanation:

The equation of a hyperbola centered in the origin in standard form is:

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

The distance between both vertexes is equal to:

`2\cdot b = \sqrt{(0-0)^(2)+(√(12)+√(12))^(2)}`

`2\cdot b = 2\cdot √(12)`

`b = √(12)`

Now, the distance between any of the vertexes and origin is:

`c = \sqrt{(0-0)^(2)+[(4-(-4)]^(2)}`

`c = 8`

The remaining parameter of the hyperbola is determined by the following Pythagorean expression:

`c^(2) = a^(2) - b^(2)`

`a = \sqrt{c^(2)+b^(2)}`

`a = √(64+12)`

`a = √(76)`

The equation of the hyperbola is:

`(x^(2))/(76) - (y^(2))/(12) = 1`

Answer 2

The equation of the hyperbola is:

x²/76 - y²/12 = 1

Explanation:

The standard for of an equation of a hyperbola centered in the origin is given as:

x²/a² - y²/b² = 1

The distance between both vertexes is:

2b, where b = √12

The distance between any of the vertexes and origin is:

c = 8

But a² = b² + c² (Pythagoras rule)

c² = a² - b²

8² = a² - 12

a² = 64 + 12 = 76

a = √76

Therefore, the equation of the hyperbola is:

x²/76 - y²/12 = 1