Question

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

Answer 1

B

Explanation:

Answer 2

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