Question

The line $2x+y=1$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ If this line passes through the point of intersection of the nearest directrix and the X-axis, then find the eccentricity of the hyperbola.

Answer 1

Answer: 2

Point of intersection is $\left(\frac{a}{e},0\right)$
Since $2x+y=1$ passes through $\left(\frac{a}{e},0\right)$
$\Rightarrow \frac{2a}{e}+0=1$
$\Rightarrow a=\frac{e}{2}$
Also $2x+y=1$
is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\Rightarrow 1^2=a^2(-2{)}^2-b^2$
$\Rightarrow 4a^2-b^2=1$
$\Rightarrow 4a^2-a^2\left(e^2-1\right)=1\dots \left[\text{ As }{b}^2=a^2\left(e^2-1\right)\right]$
$\Rightarrow 4{\left(\frac{e}{2}\right)}^2-{\left(\frac{e}{2}\right)}^2\left(e^2-1\right)=1$
$\Rightarrow e^2-\frac{e^4-e^2}{4}=1$
$\Rightarrow 5e^2-e^4=4$
$\Rightarrow e^4-5e^2+4=0$
$\Rightarrow \left(e^2-1\right)\left(e^2-4\right)=0$
$\Rightarrow e^2=4$
$\Rightarrow e=2$