The line `2x+y=1` is tangent to the hyperbla `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. If this line passes through the point of intersection of the nearest directrix and the `x`-axis, then the eccentricity of the hyperbola is

Options

A. `sqrt(2)`
B. `2`
C. `sqrt(3)`
D. `1`

Answer

B

Explanation

It is given that `2x+y=1` passes through the point `(a//e,0)`.
`:. (2a)/(e)=1impliesa=(e)/(2)`
Since `2x+y=1` i.e. `y=-2x+1` touches the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`.
`:.1^(2)=a^(2)(-2)^(2)-b^(2)` [Using : `c^(2)=a^(2)m^(2)-b^(2)`]
`implies4a^(2)-b^(2)=1`
`implies4a^(2)-a^(2)(e^(2)-1)=1` `[ :' b^(2)=a^(2)(e^(2)-1)]`
`impliese^(2)-(e^(2))/(4)(e^(2)-1)=1` `[:'a=(e)/(2)]`
`implies5e^(2)-e^(4)=4`
`impliese^(4)-5e^(2)+4=0implies(e^(2)-1)(e^(2)-4)=0impliese=2`