Question 721008
write the equation of a hyperbola in standard form whose center is (-2, -4), a focus at (-2, 6), and eccentricity of 5/4.
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Given hyperbola has a vertical transverse axis. (From given center and focus data, y-coordinates change but x-coordinates do not.)
Its standard form of equation: {{{(y-k)^2/a^2-(x-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center,
For given hyperbola:
center: (-2,-4)
c=10 (-4 to 6) (distance from center to focus on vertical transverse axis)
c^2=100
eccentricity=5/4=c/a
a=4c/5=40/5=9
a^2=81
c^2=a^2+b^2
b^2=c^2-a^2=100-81=19
Equation of given hyperbola:
{{{(y+4)^2/81-(x+2)^2/19=1}}}