Find the standard form of the equation of the hyperbola satisfying the given conditions: Foci: (0, -8) (0, 8); Vertices: (0, -6) (0, 6)

Solution:

When the hyperbola has foci on the y-axis then the equation of the hyperbola can be written as :

\(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} =1\)

Center - to - focus distance ‘c’ = \(\sqrt{a^{2} + b^{2}}\)

Foci:  (0, ±c)

Vertices: (0, ±a)

For the given problem :

The Focii are (0, ±8) and 

The vertices are (0, ±6)

Hence we can state that

c = 8 and 

a = 6

Therefore b is equal to:

b =  √(c² - a²)

b =√(8² - 6²)

b = √(64 - 36)

b = √28

b = 2√7

Hence the equation of the hyperbola is 

\(\frac{y^{2}}{36} - \frac{x^{2}}{28} = 1\)

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Find the standard form of the equation of the hyperbola satisfying the given conditions: Foci: (0, -8) (0, 8); Vertices: (0, -6) (0, 6)

Summary:

The standard form of the equation of the hyperbola satisfying the given conditions: Foci: (0, -8) (0, 8); Vertices: (0, -6) (0, 6) is \(\frac{y^{2}}{36} - \frac{x^{2}}{28} = 1\)