Question #8080a

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Answer 1 · 2017-10-20T08:46:22

See the explanation below

The equation of the hyperbola is

#x^2/a^2-y^2/b^2=1#

Comparing this equation to your equation

#25x^2-16y^2-1=0#

Rearranging your equation

#25x^2-16y^2=1#

#x^2/(1/5)^2-y^2/(1/4)^2=1#

Therefore,

#a=1/5#

#b=1/4#

#c=+-sqrt(a^2+b^2)=+-sqrt(1/25+1/16)=+-sqrt(41)/20#

The center of the hyperbola is #C=(0,0)#

The vertices are #A=(1/5,0)# and #A'=(-1/5,0)#

The foci are #F=(sqrt41/20,0)# and #F'=(-sqrt41/20,0)#

graph{25x^2-16y^2-1=0 [-1.706, 1.712, -0.853, 0.855]}