How do you find the standard form given vertices (-6, -2) and (6,-2) and foci of (-8, -2) and (8, -2)?

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Answer 1 · 2017-04-08T19:30:28

There are two standard forms for hyperbolas:
#(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"#
#(y-k)^2/a^2-(x-h)^2/b^2=1" [2]"#
Where #(h,k)# is the center point.

Because the vertices have the same y coordinate but different x coordinates, we know that we must use equation [1].

For equation [1] here are the general formulas for the vertices are:

#x_1 = h - a#
#x_2 = h + a#
#y_1=y_2 = k#

Substitute -6 for #x_1#, 6 for #x_2# and -2 for #y_1#:

#-6 = h - a#
#6 = h + a#
#-2 = k#

It is obvious that #h = 0#, #k = -2#, and #a = 6#.

Substitute these values into equation [1]:

#(x-0)^2/6^2-(y--2)^2/b^2=1" [3]"#

For equation [1] here are the general formulas for the foci are:

#x_1 = h - sqrt(a^2+b^2)#
#x_2 = h + sqrt(a^2+b^2)#
#y_1 = y_2 = k#

Substitute 8 for #x_2#, 0 for h and 6 for a:

#8 = 0 + sqrt(6^2+b^2)#

Solve for b:

#64 = 36+b^2#

#b^2 = 28#

#b = sqrt(28)#

Substitute this into equation [3]:

#(x-0)^2/6^2-(y--2)^2/(sqrt(28))^2=1" [4]"#

Equation [4] is the answer.