Q98 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics

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

  • A
    $\sqrt{2}$
  • Correct answer
    B
    2
  • C
    $\sqrt{3}$
  • D
    3

Solution

The line $2 x+y-1=0$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and also passes through the point of intersection of directrix and the positive $x$-axis Equation of direction $x=\frac{a}{e}$ $\therefore$ Point $\left(\frac{a}{e}, 0\right)$ passes through $2 x+y=1$ $\Rightarrow \quad 2\left(\frac{a}{e}\right)+0=1$ $\Rightarrow \quad 2 a=e$ $\Rightarrow \quad 2 a^2=a e$ We know, $(a e)^2=a^2+b^2$ $\therefore \quad 4 a^4=a^2+b^2$ Equation of tangent of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $y=m x \pm \sqrt{a^2 m^2-b^2}$ $\therefore \quad a^2(2)^2-b^2=1$ $$ \Rightarrow \quad 4 a^2-b^2=1 $$ From Eqs. (i) and (ii), we get $$ \begin{array}{ll} & 4 a^4=a^2+4 a^2-1 \\ \Rightarrow & 4 a^4-5 a^2+1=0 \\ & \left(4 a^2-1\right)\left(a^2-1\right)=0 \\ \Rightarrow & a^2=1, \frac{1}{4} \Rightarrow a=1, \frac{1}{2} \\ \therefore & e=2 a \\ & e=2(1)=2 \end{array} $$

Hence, the correct answer is option: 'B'.

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