Question

Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1 divided by 4.x

Answer 1

Answer: y^2/16-x^2/256=1

Solution:

The vertices are (0,±4): (0,-4) and (0,4)

The parabola has a vertical axis (the line that joins the vertices).

The center C=(h-k) is the midpoint between the vertices, then:

h=(0+0)/2→h=0/2→h=0

k=(-4+4)/2→k=0/2→k=0

C=(h,k)→C=(0,0)

The equation of the parabola has the form:

y^2/a^2-x^2/b^2=1

The distance between the vertices and the center is a, then a=4

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

y^2/16-x^2/b^2=1 (1)

The asympotes of this parabola are:

y=±(a/b)x

The given asympotes are: y=±(1/4)x, then:

a/b=1/4; a=4

4/b=1/4

Solving for b. Cross multiplication:

(4)(4)=(1)(b)

16=b

b=16

Replacing in the equation (1)

(1) y^2/16-x^2/16^2=1

y^2/16-x^2/256=1

Equation of the hyperbola:

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

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

y^2-x^2/16=1