Question

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

Answer 1

The hyperbola is a vertical hyperbola which means the equation is (y^2/a^2)-(x^2/b^2)

The vertices is at (0+/-2) meaning a=2

so we have (y^2/2^2)-(x^2/b^2)=1

The foci is at (0+/-7) meaning b=7

so now we have (y^2/2^2)-(x^2/7^2)=1

Then we simplify and the equation is (y^2/4)-(x^2/49)=1

Answer 2

Standard Form of an Equation of an Hyperbola opening up and down is:

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

where Pt(h,k) is a center with vertices 'b' units up and down from center.

vertices: vertices: (0,+-2) b = 2 AND Center is (0,-0)

(y)^2/4-(x)^2/a^2=1

foci: (0,+-7) 

a^2 = 24 

square root of 4 + a^2 = 5.29

4 + 5.29 = 9.29

y^2 / 4 - x^2 / 24 = 1