How do you write an equation for a hyperbola with vertices (1, 3) and (-5, 3), and foci (3, 3) and (-7, 3)?

1 Answer
Oct 30, 2016

Answer:

The equation is:

#(x - -2)^2/3^2 - (y - 3)^2/4^2 = 1#

Explanation:

Please notice that the vertices are of the forms:

#(h - a, k)# and #(h + a,k)# specifically #(-5,3)# and #(1, 3)#

The same information can be deduced from the foci, which have the forms:

#(h - c, k)# and #(h + c, k)# specifically #(-7, 3)# and #(3, 3)#

The standard form for the equation of a hyperbola, where the vertices and foci have these properties, is the horizontal transverse axis form:

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

#k = 3# by observation:

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

Compute h and a:

#-5 = h - a# and #1 = h + a#

#2h = -4#

#h = -2#

#a = 3#

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

To complete the equation, we only need the value of b but, to find the value of b, we must, first, find the value of c:

Using the #(h + c, k)# form for the focus point, #(3,3)#, we substitute -2 for h, set the right side equal to 3, and then solve for c:

#-2 + c = 3#

#c = 5#

Solve for b, using the equation #c^2 = a^2 + b^2#:

#5^2 = 3^2 + b^2#

#b = 4#

#(x - -2)^2/3^2 - (y - 3)^2/4^2 = 1#