Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11).

Solution:

Vertices at (0, ±9) and foci at (0, ±11) [Given]

As both vertices and foci lie on the Y-axis, the equation of the hyperbola is

y²/b² - x²/a² = 1

Vertices = (0, ±9)

b = 9

Foci = (0, ±11)

be = 11

e = 11/9

We know that

e = √(1 + a²/b²)

Substituting the values

11/9 = √(1 + a²/b²)

Squaring on both sides

121/81 = 1 + a²/b²

a²/b² = 121/81 - 1

a²/b² = 40/81

Here

a² = 40/81 × 9²

a² = 40/81 × 81

a² = 40

a = 2√10

So the equation of hyperbola is

y²/81 - x²/40 = 1

Therefore, the equation of the hyperbola in standard form is y²/81 - x²/40 = 1.

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Find an equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11).

Summary:

An equation in standard form for the hyperbola with vertices at (0, ±9) and foci at (0, ±11) is y²/81 - x²/40 = 1.