Question 359583
I think you mean plot the function. 
It's difficult without having some handle on {{{a}}} and {{{b}}}.
The function would be of the form,
{{{y^2-b^2=a^2b^2/(x^2-a^2)}}}
{{{y^2=a^2b^2/(x^2-a^2)+b^2}}}
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Here's the effect of changing {{{b}}}, keeping {{{a}}} fixed.
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{{{drawing(300,300,-5,5,-5,5,grid(1),
graph(300,300,-5,5,-5,5,x/sqrt(x^2-1^2),2x/sqrt(x^2-1^2),3x/sqrt(x^2-1^2)),
graph(300,300,-5,5,-5,5,-x/sqrt(x^2-1^2),-2x/sqrt(x^2-1^2),-3x/sqrt(x^2-1^2)
))}}}
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.Red:{{{a=1}}},{{{b=1}}}
Green:{{{a=1}}},{{{b=2}}}
Blue:{{{a=1}}},{{{b=3}}}
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Here's the effect of changing {{{a}}}, keeping {{{b}}} fixed.
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{{{drawing(300,300,-5,5,-5,5,grid(1),
graph(300,300,-5,5,-5,5,x/sqrt(x^2-1^2),x/sqrt(x^2-2^2),x/sqrt(x^2-3^2)),
graph(300,300,-5,5,-5,5,-x/sqrt(x^2-1^2),-x/sqrt(x^2-2^2),-x/sqrt(x^2-3^2)
))}}}
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Red:{{{a=1}}},{{{b=1}}}
Green:{{{a=2}}},{{{b=1}}}
Blue:{{{a=3}}},{{{b=1}}}
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(0,0) is a solution independent of {{{a}}} and {{{b}}}
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When {{{a=0}}}, the solutions are of the form, {{{y = 0 +- b}}} and {{{x=0}}}.
When {{{b=0}}}, the solutions are of the form, {{{x = 0 +- a}}} and {{{y=0}}}.
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