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

Asked by maham237 @ 22/12/2020 in Mathematics viewed by 323 People

Answered by maham237 @ 22/12/2020

Answer:

$\frac{y^2}{4}-\frac{x^2}{117}=1$

Step-by-step explanation:

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

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ ....[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;

$\frac{y^2}{4}-\frac{x^2}{117}=1$

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

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