Question

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 $\frac{x^2}{a^2}-\frac{y^2}{4}=1,$ then eccentricity of hyperbola is

Answer 1

$a+c=2b$
$\Rightarrow a-2b+c=0$
So, given family of lines $ax+by+c=0$ will always pass through $(1,-2)$
Since, $(1,-2)$ lies on the hyperbola also
$\therefore$ Putting in $\frac{x^2}{a^2}-\frac{y^2}{4}=1$, we get
$\frac{1}{a^2}-\frac{4}{4}=1$
$\Rightarrow \frac{1}{a^2}=2\Rightarrow a^2=\frac{1}{2}$
$\therefore e^2=1+\frac{b^2}{a^2}=1+4\times 2$
$\Rightarrow e^2=9$
$\therefore e=3$