NCERT Solutions
Class 11 Maths
Conic Sections

Ex.11.4 Q.15
Find the equation of the hyperbola satisfying the give conditions: Foci (0, ±√10), passing through (2, 3).
Given, Foci (0, ±√10), passing through (2, 3).
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form (y2 ÷ a2) – (x2 ÷ b2) = 1.
Since the foci are (0, ±√10), c = √10
We know that a2 + b2 = c2
=> a2 + b2 = 10
=> b2 = 10 - a2 …………… (1)
Since the hyperbola passes through point (2, 3),
(9 ÷ a2) – (4 ÷ b2) = 1 ………... (2)
From equations (1) and (2), we obtain
(9 ÷ a2) – {4 ÷ (10 - a2)} = 1
=> 9(10 - a2) - 4a2 = a2 (10 - a2)
=> 900 - 9a2 - 4a2 = 100a2 – a4
=> a4 – 23a2 + 90 = 0
=> a4 – 18a2 - 5a2 + 90 = 0
=> a2 (a2 - 18) - 5(a2 - 18) = 0
=> (a2 - 18) (a2 - 5) = 0
=> a2 = 18, 5
In hyperbola, c > a,
=> c2 > a2
So, a2 = 5
Now, b2 = 10 – a2
=> b2 = 10 – 5
=> b2 = 5
Thus, the equation of the hyperbola is y2 -
x2 = 1.