Question

The eccentricity of a hyperbola is $\sqrt{2}$ and its centre is $(1,1)$ if the equation of one asymptote is $2x+3y-5=0$ then the equation of the other asymptote is :

• A. $3x-2y=0$
• B. $3x-2y=1$
• C. $3x-2y+1=0$
• D. $3x-2y+3=0$

Answer 1

The correct option is B $3x-2y=1$

Given,
Eccentricity of a hyperbola is $\sqrt{2}$ and its centre is (1, 1)
Equation of one Asymptote is 2x+ 3y - 5 = 0
Since Both the Asymptote are perpendicular to each other
Therefore ,
Equation of a line which is Perpendicular to 2x+ 3y - 5 = 0
$3x-2y-\lambda =0$
Since , it Passes through (1, 1) So
$3-2-\lambda =0$
$\lambda =1$
Finally , we get
$3x-2y-1=0$
$3x-2y=1$
Hence , equation of the other asymptote is
$3x-2y=1$