IIT-JEE 2010 · Paper 1, Mathematics question 50

Directrix point. The nearest directrix $x=\frac ae$ meets the $x$-axis at $\left(\frac ae,0\right)$; it lies on $2x+y=1$, so $a=\frac e2$.

Tangency. $y=-2x+1$ touches the hyperbola when $1=4a^2-b^2=a^2(5-e^2)$ (using $b^2=a^2(e^2-1)$):
$$1=\frac{e^2}4(5-e^2)\implies e^4-5e^2+4=0\implies e^2=4\text{ or }1.$$
A hyperbola has $e>1$, so $e=2$.

Answer: 2