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, a2 + b2 = c2
b2 = 10 – a2 ………….. (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 – a2) – 4a2 = a2(10 –a2)
90 – 9a2 – 4a2 = 10a2 – a4
a4 – 23a2 + 90 = 0
a4 – 18a2 – 5a2 + 90 = 0
a2(a2 -18) -5(a2 -18) = 0
(a2 – 18) (a2 -5) = 0
a2 = 18 or 5
In hyperbola, c > a i.e., c2 > a2
So, a2 = 5
b2 = 10 – a2
= 10 – 5
= 5
The equation of the hyperbola is
\( \dfrac{y^2}{5}- \dfrac{x^2}{5}=1\)
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