Question

Find an equation in standard form for the hyperbola with vertices at (0, ±2) and foci at (0, ±7).

Answer 1

Explanation:

std form: (y - k)^2 / a^2 - (x - h)^2 / b^2 = 1

vertices at (0, ±2) so x=0, y=±2

by symmetry, h=0 n k=0

(±2)^2 / a^2 = 1

a^2 = 4

using formula c^2 = a^2 + b^2 where c=foci at (0, ±7)

(±7)^2 = 4 + b^2

b^2 = 49 - 4 = 45

put everything back in

y^2/4 - x^2/45 = 1

Answer 2

The equation in standard form is y^2/4 - x^2/45 = 1

Explanation:

As vertices at (0, ±2), the hyperbola opens up and down,

the standard form for a hyperbola that opens up and down is:

(y - k)^2 / a^2 - (x - h)^2 / b^2 = 1,

where its center is located at (h, k);

a is 2x distance between the vertices;

c is 2x distance between the foci;

and b^2 = c^2 - a^2

As vertices are at (0, ±2) and foci are at (0, ±7),

a=2 and c=7

a^2 = 2^2 = 4

b^2 = c^2 - a^2

= 7^2 - 2^2

= 45

The center is at the mid-pt between vertices,

so it is at (0, 0)

h=0 and k=0

Substituting a^2, b^2, h and k in the standard form,

y^2/4 - x^2/45 = 1