Foci (0, ±\( \sqrt{10}\)) and passing through (2, 3)

Here, the foci are on y-axis.

The equation of the hyperbola is of the form

\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1

Since, the foci are (±\( \sqrt{10}\), 0), so, c =\( \sqrt{10}\)

It is known that, a^{2} + b^{2} = c^{2}

b^{2} = 10 – a^{2} ………….. (1)

It is given that the hyperbola passes through point (2, 3)

So, \( \dfrac{9}{a^2}\) – \( \dfrac{4}{b^2}\) = 1 … (2)

From equations (1) and (2), we get,

\( \dfrac{9}{a^2}-\dfrac{10}{a^2}=1\)

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^{4} – 18a^{2} – 5a^{2 }+ 90 = 0

a^{2}(a^{2} -18) -5(a^{2} -18) = 0

(a^{2} – 18) (a^{2} -5) = 0

a^{2} = 18 or 5

In hyperbola, c > a i.e., c^{2 }> a^{2}

So, a^{2} = 5

b^{2} = 10 – a^{2}

= 10 – 5

= 5

The equation of the hyperbola is

\( \dfrac{y^2}{5}- \dfrac{x^2}{5}=1\)

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