If a,b,c are in A.P, the straight line ax+by+c=0, passes through a fixed point which lie on the hyperbola x2a2−y24=1, then eccentricity of hyperbola is
Open in App
Solution
a+c=2b ⇒a−2b+c=0
So, given family of lines ax+by+c=0 will always pass through (1,−2)
Since, (1,−2) lies on the hyperbola also ∴ Putting in x2a2−y24=1, we get 1a2−44=1 ⇒1a2=2⇒a2=12 ∴e2=1+b2a2=1+4×2 ⇒e2=9 ∴e=3