Question

help with this?

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Answer 1

yes first tell me which type of conic this is?

Answer 2

its a hyperbola it'nt it?

Answer 3

ya

Answer 4

ok standard form for this type of hyperbola
\[\frac{ (x-\alpha)^2}{ a^2 }-\frac{ (y-\beta)^2 }{ b^2 }=1\\for~this~model--\\center=(\alpha,\beta)\\vertices=(\alpha \pm a,\beta)\\elccentricity=e=\sqrt{1+\frac{ b^2 }{ a^2 }}\\focii=(\alpha \pm ae,\beta)\]
thats all u need now just algebra :)