You can put this solution on YOUR website! Write standard form of hyperbola.
Foci: (+-10,0) asymptotes: y=(+-3/4)x
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Given hyperbola has a horizontal transverse axis with center at the origin. (gleaned from foci data)
Its standard form of equation:
..
For given hyperbola:
center:(0,0)
slope of asymptotes=±3/4=b/a
b=3a/4
c=10(distance from center to foci)
c^2=a^2+b^2
c^2=a^2+(3a/4)^2
100=a^2+9a^2/16
1600=16a^2+9a^2
1600=25a^2
a^2=1600/25=64
b^2=9a^2/16
b^2=9*64/16=36
Equation of given hyperbola:
(