How do you write an equation for the hyperbola with a centre at (-3,1), one focus (-5,1), length of conjugate axis 4?

1 Answer
Oct 16, 2016

Answer:

One cannot have a hyperbola with the equation:

#(x - -3)^2/0^2 - (y - 1)^2/2^2 = 1#

and that is what your specifications imply.

Explanation:

Because the center and the given focus share the same y coordinate, we know that this is a horizontal transverse type with a general equation of the form:

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

We are given that the center #(h, k)# is #(-3, 1)# so we substitute these values into the equation:

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

We are given the length of the conjugate axis is 4; this allows us to find b:

#2b = 4#

#b = 2#

Substitute this into the equation:

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

We are given that a focus is at #(-5, 1)# and we know that the foci are at (h - c, k) and (h + c, k). Because c is always positive, we can only obtain a number more negative than the center's x coordinate, using the first focus:

#-5 = h - c#

#-5 = -3 - c#

#-2 = -c#

#c = 2#

We can find #a# using the equation:

#c^2 = a^2 + b^2#

#2^2 = a^2 + 2^2#

#a^2 = 0#

#a = 0#