Question 1136096


Find an equation in standard form for the hyperbola with 
vertices at ({{{0}}}, ±{{{9}}}) and 
foci at ({{{0}}}, ±{{{11}}}). 

The standard form of the equation for this type of hyperbola given the center is at the origin is:
 
{{{y^2 / a^2 - x^2 / b^2 =1 }}}

Your coordinates of the vertices are ({{{0}}}, ±{{{9}}}) and ({{{0}}}, ±{{{11}}}) making {{{a=9}}} and {{{c=11}}}


{{{c^2=a^2+b^2}}}

{{{11^2=9^2+b^2}}}

{{{b^2=121-81}}}

{{{b^2=40}}}

so your equation is:

{{{y^2 / 81 - x^2 / 40 =1 }}}


answer: C.y squared over eighty one minus x squared over forty = 1