Q84 / 150 · MHT CET 2008 · Mathematics

The equation of the tangent parallel to $y-x+5=0$ drawn to $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ is

  • Correct answer
    A
    $x-y-1=0$
  • B
    $x-y+2=0$
  • C
    $x+y-1=0$
  • D
    $x+y+2=0$

Solution

Given hyperbola is $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ ...(i) Equation of tangent parallel to $y-x+5=0$ is $$ \begin{aligned} & & y-x+\lambda &=0 & \\ \Rightarrow & & y &=x-\lambda ...(ii) \end{aligned} $$ If line (ii) is a tangent to hyperbola (i), then $$ \begin{array}{rc} -\lambda=\pm \sqrt{3 x-2} \\ \left(\text { from } c=\pm \sqrt{a^{2} m^{2}-b^{2}}\right) \\ \Rightarrow & -\lambda=\pm 1 \\ \Rightarrow & \lambda=-1,+1 \end{array} $$ Put the values of $\lambda$ in Eq. (ii), we get $x-y-1=0 \quad$ and $\quad x-y+1=0$ are the required tangents.

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

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