Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±10).

y squared over 81 minus x squared over 100 = 1

y squared over 81 minus x squared over 19 = 1

y squared over 19 minus x squared over 81 = 1

y squared over 100 minus x squared over 81 = 1

Respuesta :

Answer:

y² / 81  -  x² / 19   = 1

Step-by-step explanation: See Annex ( vertices and foci in coordinates axis)

The equation in standard form for the hyperbola is:

x² / a² - y²/b² = 1       or    y²/a²  -  x² / b²  = 1

In cases of transverse axis parallel to x axis  or y axis respectively.

As per given information in this case hyperbola has a transverse axis parallel to  y  axis the equation is

y²/a²  -  x² / b²  = 1

a is a distance between  center and vertex therefore a = 9

c is a distance between center and a focus c = 10

and  b will be:

c² = a²  +  b²    ⇒   b²  = c²  - a²     ⇒ b²  =  (10)² - (9)²    ⇒  b² = 100 - 81

b = √19

And the equation in standard form is:

y² / a²  -  x² / b²  = 1

y² / ( 9 )²   -  x² / √(19)²     ⇒   y² / 81  -  x² / 19   = 1

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Answer:

B

Step-by-step explanation: