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

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