JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Hyperbola

The nearest directrix of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x=\frac{a}{e}$, where $e$ is the eccentricity. Its intersection with the x-axis is $(\frac{a}{e},0)$.

The line $2x+y=1$ passes through this point: $2\cdot \frac{a}{e}+0=1$, giving $a=\frac{e}{2}$.

For the line $y=-2x+1$ to be tangent to the hyperbola, substitute into the hyperbola equation:


Expanding and rearranging: $(b^2-4a^2)x^2+4a^{2}x-a^2(b^2+1)=0$.

For tangency, discriminant = 0: $16a^4+4a^2(b^2-4a^2)(b^2+1)=0$.

Using $b^2=a^2(e^2-1)$ and $a=\frac{e}{2}$, substitute and solve: $e^2-1=3$, giving $e^2=4$, so $e=2$.