Question 1153742

given:

foci ({{{0}}},{{{13}}}), ({{{0}}},{{{-13}}})

vertices ({{{0}}},{{{5}}}),  ({{{0}}},{{{-5}}})

We can tell that it is a vrtical hyperbola.

The center point is ({{{0}}}, {{{0}}}). 

To find {{{a}}}, we'll count from the center to either vertex. 

{{{a = 5}}}

To find {{{c}}}, we'll count from the center to either focus. 

{{{c = 13}}}

then use

{{{c^2 = a^2 + b^2}}}
{{{b^2 = c^2 -a^2}}}
{{{b^2 = 13^2 -5^2}}}
{{{b^2 = 169 -25}}}
{{{b^2 = 144}}}
{{{ b=12}}}

We have all our information:{{{ h = 0}}}, {{{k = 0}}}, {{{a = 5}}}, {{{b = 12}}}. Since it's a horizontal hyperbola centered in origin, we'll choose that formula and substitute in our information.

{{{y^2/a^2-x^2/b^2=1}}}

{{{y^2/5^2-x^2/12^2=1}}}

{{{y^2/25-x^2/144=1}}}


{{{drawing(600, 600, -15, 15, -15, 15, 
circle(0,13,.17),circle(0,-13,.17),circle(0,5,.17),circle(0,5,.17),
locate(0.3,13.5,F(0,13)),locate(0.3,-13.5,F(0,-13)),
locate(0.5,5,v(0,5)),locate(0.5,-5,v(0,-5)),
 graph( 600, 600, -15, 15, -15, 15, -sqrt(25x^2/144+25), sqrt(25x^2/144+25))) }}}