Question 716101
find an equation of the hyperbola with vertices at (-6, 1) and (4, 1) and foci at (-8, 1) and (6, 1)
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Standard form of a hyperbola with horizontal transverse axis:
{{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of vertex.
y-coordinate of center=1
x-coordinate of center=(-6+4)/2=-1 (midpoint formula)
center: (-1, 1)
a=5 (distance from center to vertex, (-1 to - 6) on the horizontal transverse axis)
a^2=25
c=7 (distance from center to focus, (-1 to - 8) on the horizontal transverse axis)
c^2=49
c^2=a^2+b^2
b^2=c^2-a^2=49-25=24
Equation of given hyperbola:
{{{(x+1)^2/25-(y-1)^2/24=1}}}