Given hyperbola is 9x2−b2y2=1 Since this passes through (K,2), therefore 9K2−b24=1 Also, given e=1+a2b2=313 ⇒1+9b2=313⇒9+b2=13⇒b=±2 Now, from eqn (1), we have 9K2−44=1(∵b=±2)⇒K2=18
If the eccentricity of a hyperbola x^2 9- y^2 b^2=1, which passes through (K, 2), is √13 3, then the value of K^2 is
JEE Main 2012 — Mathematics Coordinate Geometry
If the eccentricity of a hyperbola 9x2−b2y2=1, which passes through (K,2), is 313, then the value of K2 is
Held on 7 May 2012 · Verified 6 Jul 2026.
Options
- A
18
- B
8
- C
1
- D
2
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