How do you write an equation for the hyperbola with focus at (0, 25), asymptotes intersecting at (0, 5) and an asymptote passing through (24, 13)?

1 Answer
Aug 31, 2016

Answer:

#(y-5)^2/64-x^2/336=1#

Explanation:

The asymptotes intersect at the center #C(0., 5)#.

A focus is #S(0, 25)#.

#CS = ae = 25-5= 20#.,

where a is the semi transverse axis and e is the eccentricity.

The transverse axis is along the y-axis. So, the equation of the

hyperbola is

#(y-5)^2/a^2-(x-0)^2/(a^2(1-e^2))=1#

As it passes through (24, 13),

#8^2/a^2-24^2/(a^2(e^2-1))=1#

Substituting a = 20/e and simplifying,

#4e^4-29e^2+25=0#

#4e^4-25e^2-4e^2+25=0#

#e^2(4e^2-25)-(4e^2-25)=0#

#(4e^2-25)(e^2-1)=0#

As e > 1,

#4e^2-25=0#

#e=5/2#.

#a=20/e=20/(5/2)=8#
^Semi conjugate axis #b= a sqrt(e^2-1)=8sqrt (21/4) and b^2=336#

And now, the equation is obtained as

#(y-5)^2/64-x^2/336=1#-.