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(A) The foci are $(0, \pm \sqrt{10})$,which lie on the $y$-axis. Thus,the equation is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$. Her

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$\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1$

Vedclass · Answer

(A) The foci are $(0, \pm \sqrt{10})$,which lie on the $y$-axis. Thus,the equation is of the form $\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$. Here,$c = \sqrt{10}$,so $c^{2} = 10$. We know $c^{2} = a^{2} + b^{2}$,so $a^{2} + b^{2} = 10$,which means $b^{2} = 10 - a^{2}$. The hyperbola passes through $(2, 3)$,so $\frac{3^{2}}{a^{2}} - \frac{2^{2}}{b^{2}} = 1$,or $\frac{9}{a^{2}} - \frac{4}{b^{2}} = 1$. Substituting $b^{2} = 10 - a^{2}$: $\frac{9}{a^{2}} - \frac{4}{10 - a^{2}} = 1$. $9(10 - a^{2}) - 4a^{2} = a^{2}(10 - a^{2})$. $90 - 9a^{2} - 4a^{2} = 10a^{2} - a^{4}$. $a^{4} - 23a^{2} + 90 = 0$. $(a^{2} - 18)(a^{2} - 5) = 0$. Since $c^{2} > a^{2}$,$10 > a^{2}$,so $a^{2} = 5$. Then $b^{2} = 10 - 5 = 5$. The equation is $\frac{y^{2}}{5} - \frac{x^{2}}{5} = 1$.

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm \sqrt{10})$,passing through $(2, 3)$.

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