How do you find the foci and sketch the hyperbola #9y^2-x^2=4#?

1 Answer
Dec 27, 2016

Please read the explanation.

Explanation:

The standard form for a hyperbola of this type is:

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

where, x and y correspond to any point #(x,y)# on the hyperbola, h and k correspond to the center point, #(h,k)#, "a" is the vertical distance from the center to the vertices, and #sqrt(a^2 + b^2)# is the vertical distance from the center to the foci.

Given: #9y^2 - x^2 = 4#

We will be using algebraic steps to put the above equation into the standard form.

Divide both sides by 4 in the form of #2^2#:

#9y^2/2^2 - x^2/2^2 = 1#

Write the 9 as #3^2# and the denominator of the divisor:

#y^2/(2^2/3^2) - x^2/2^2 = 1#

Group the denominator of the first term into a single fraction:

#y^2/((2/3)^2) - x^2/2^2 = 1#

#sqrt(a^2 + b^2) = sqrt((2/3)^2 + 2^2) = (2sqrt(10))/3#

The foci are located at #(0, -(2sqrt(10))/3) and (0, (2sqrt(10))/3)#

To help you sketch the hyperbola, the center is #(0,0)# and the vertices are at #(0,-2/3) and (0,2/3)#

There is a graph of the original equation:

Desmos.com