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)?

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Answer 1 · 2016-08-31T16:27:19

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

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

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