Question

Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).

Answer 1

Answer:

$\dfrac{y^2}{36}-\dfrac{x^2}{45}=1.$

Step-by-step explanation:

Since vertices and foci lie on the y-axis, the equation of the hyperbola is

$\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1.$

If the vertices are at points (0,±6), then $b=6.$

If the foci are at points (0,±9), then $c=9.$

Note that

$c^2=b^2+a^2,$

then

$9^2=6^2+a^2,\\ \\a^2=81-36,\\ \\a^2=45.$

The equation of the hyperbola is

$\dfrac{y^2}{36}-\dfrac{x^2}{45}=1.$