Question 715112
use the eccentricity of each hyperbola to find its equation in standard form eccentricity 4 , vertices (-1,3) and (-1,7)
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Standard form of equation for a hyperbola with vertical transverse axis:
{{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center
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x-coordinate of center=-1
y-coordinate of center= (3+7)/2=5 (midpoint formula)
center:(-1,5)
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length of vertical transverse axis=4 (3 to 7)=2a
a=2
a^2=4
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c=distance from center to foci
Eccentricity=4=c/a
c=4a
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c^2=a^2+b^2
16a^2=a^2+b^2
b^2=16a^2-a^2=15a^2
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equation of given hyperbola:
{{{(y-5)^2/a^2-(x+1)^2/b^2=1}}}
{{{(y-5)^2/a^2-(x+1)^2/15a^2=1}}}
{{{(y-5)^2/4-(x+1)^2/60=1}}}