Question

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

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

Answer 1

Answer: (A)

$x-y-1=0$

Given hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$ .....(i)
Equation of tangent parallel to $y-x+5=0$ is
$y-x+\lambda =0$
$\Rightarrow y=x-\lambda$ .....(ii)
If line (ii) is a tangent to hyperbola (i), then
$-\lambda =\pm \sqrt{3x-2}$
$(fromc=\pm \sqrt{a^2{m}^2-b^2})$
$\Rightarrow -\lambda =\pm 1$
$\Rightarrow \lambda =-1,+1$
Put the values of $\lambda$ in equation (ii), we get $x-y-1=0$ and $x-y+1=0$ are the required tangents.