Question

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11).

Answer 1

`(y^2)/(4)-(x^2)/(117)=1`

Explanation:

The standard form of hyperbola centered at origin is given by:

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

where,

vertices and foci are
`(0, \pm a)` and
`(0, \pm c)` respectively.

As per the statement:

The hyperbola with vertices at (0, ±2) and foci at (0, ±11).

⇒a = 2 and c = 11

⇒
`a^2 = 4` and
`c^2 = 121`

To find
`b^2`:

Using the equation:

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

then;

`b^2 = 11^2-2^2 = 121-4 = 117`

⇒
`b^2 = 117`

Substitute the given values in [1] we have;

`(y^2)/(4)-(x^2)/(117)=1`

Therefore, an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±11) is,
`(y^2)/(4)-(x^2)/(117)=1`