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

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

chemists are often responsible for designing a chemical reaction and analyzing the products obtained from it. one significant part of designing a chemical reaction is determining the amount of each reactant that is needed and the amount of each product that will be produced. in order to do so, chemists use stoichiometry, the quantitative relationships between reactants and products.