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# Find the equation of the hyperbola with foci $(0,\pm\sqrt{10})$,passing through $(2,3)$.

$\begin{array}{1 1}\large\frac{x^2}{5}+\frac{y^2}{5}\normalsize =1\\\large\frac{x^2}{5}-\frac{y^2}{5}\normalsize =1\\\large\frac{x^2}{5}+\frac{y^2}{5}\normalsize =-1\\\large\frac{x^2}{5}-\frac{y^2}{5}\normalsize =-1\end{array}$

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• If the vertices of the hyperbola lie on the $y$-axis,then the equation of the hyperbola is $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1 where a^2=b^2(e^2-1). Answer : \large\frac{x^2}{5}-\frac{y^2}{5}\normalsize =-1 Given data is the coordinates of foci are (0,\pm\sqrt{10}) The hyperbola passes through the point (2,3). Hence the equation of the hyperbola is \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=-1$
It passes through $(2,3)$
$\therefore \large\frac{4}{a^2}-\frac{9}{b^2}$$=-1 But a^2=b^2(e^2-1) \therefore \large\frac{4}{b^2(e2-1)}-\frac{9}{b^2}$$=-1$------(1)
The coordinates of the foci are $(0,\pm \sqrt{10})$
$\therefore be=\sqrt{10}$
$\Rightarrow b^2e^2=10$
Hence $\large\frac{4}{10-b^2}-\frac{9}{b^2}$$=-1 On simplifying we get, 4b^2-9(10-b^2)=-b^2(10-b^2) \Rightarrow 4b^2-90+9b^2=-10b^2+b^4 \Rightarrow b^4-23b^2+90=0 On factorizing we get, (b^2-18)(b^2-5)=0 \therefore b^2=18 or b^2=5 a^2=b^2(e^2-1) a^2=10-b^2 \;\;\;\;=10-18=-8 which is not possible. a^2=10-5=5 Hence the required equation is \large\frac{x^2}{5}-\frac{y^2}{5}$$=-1$