If (0, pm 4) and (0, pm 2) are the foci and v | vedclass

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(C) The foci are given as $(0, \pm 4)$,which corresponds to $(0, \pm be)$. Thus,$be = 4$.The vertices are given as $(0, \pm 2)$,which corresponds to $

Vedclass · Accepted Answer

$\frac{y^2}{4} - \frac{x^2}{12} = 1$

Vedclass · Answer

(C) The foci are given as $(0, \pm 4)$,which corresponds to $(0, \pm be)$. Thus,$be = 4$.The vertices are given as $(0, \pm 2)$,which corresponds to $(0, \pm b)$. Thus,$b = 2$.Substituting $b = 2$ into $be = 4$,we get $2e = 4$,so $e = 2$.Using the relation $a^2 = b^2(e^2 - 1)$ for a vertical hyperbola,we have $a^2 = 2^2(2^2 - 1) = 4(4 - 1) = 4(3) = 12$.The standard equation for a vertical hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$.Substituting the values,we get $\frac{y^2}{4} - \frac{x^2}{12} = 1$.

If $(0, \pm 4)$ and $(0, \pm 2)$ are the foci and vertices of a hyperbola,respectively,then its equation is:

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