Find the equation of the hyperbola with foci | vedclass

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Question

(A) Since the foci are on the $y$-axis,the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.Given the vertices

Vedclass · Accepted Answer

$100 y^{2} - 44 x^{2} = 275$

Vedclass · Answer

(A) Since the foci are on the $y$-axis,the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$.Given the vertices are $(0, \pm \frac{\sqrt{11}}{2})$,we have $a = \frac{\sqrt{11}}{2}$,so $a^{2} = \frac{11}{4}$.Given the foci are $(0, \pm 3)$,we have $c = 3$,so $c^{2} = 9$.For a hyperbola,$b^{2} = c^{2} - a^{2} = 9 - \frac{11}{4} = \frac{36 - 11}{4} = \frac{25}{4}$.Substituting $a^{2}$ and $b^{2}$ into the standard equation:$\frac{y^{2}}{11/4} - \frac{x^{2}}{25/4} = 1$$\frac{4y^{2}}{11} - \frac{4x^{2}}{25} = 1$Multiplying by $275$ (the $LCM$ of $11$ and $25$):$100y^{2} - 44x^{2} = 275$.

Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $(0, \pm \frac{\sqrt{11}}{2})$.

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