Foci (0, ± ), passing through (2, 3)

Question

Philoid · Accepted Answer

Since the foci are on y-axis, the equation the hyperbola is of the form

= 1
Since foci are (0, ± ), c =

Given that point(2, 3) lies on the hyperbola, we have

= 1 ⇒ 9b2 – 4a2 = a2b2
Also, c2 = a2 + b2 or 9b2 – 4a2 = a2b2

or b2 = 10 – a2

substituting the value of b2 in

9b2 – 4a2 = a2b2 we get

or 9(10 – a2) – 4a2 = a2(10 – a)2

or 90 – 9a2 - 4a2 = 10a2 – a4

or 90 – 23a2 – a4 = 0

or 90 – 18a2 – 5a2 + a4 = 0

or 18(5 – a2)(18 – a2) = 0

From a2 + b2 = 10 we have

5 + b2 = 10 we have

5 + b2 = 10 or 18 + b2 = 10

or b2 = 5 or b2 = -8

Rejecting –ve value, we have b2 = 5 and a2 = 5. Therefore, the equation of the required hyperbola is

= 1 ⇒ y2 – x2 = 5.