Question

The equation of one asymptote of the hyperbola $14x^2+38xy+20y^2+x-7y-91=0$ is $7x+5y-3=0$ then the other asymptote is :

• A. $2x+4y=1$
• B. $2x-4y=1$
• C. $2x+4y+1=0$
• D. $2x-4y+1=0$

Answer 1

Answer: (C)

$2x+4y+1=0$

Observing from the given hyperbola, we take that the equation of the asymptotes is of the form

$14x^2+38xy+20y^2+x-7y-91=0$

Thus can be written as $14x^2+28xy+10xy+20y^2+x-7y-91=0$

or

$(14x+10y)(x+2y)+x-7y-91=0$.................1

Now let the equations be $(14x+10y+l)(x+2y+m)=0$

or

$14x^2+38xy+20y^2+(14m+l)x+(10m+2l)y+lm=0$......................2

compare coefficients in 1 and 2 to get

$(14m+l)=1$, $(10m+2l)=-7$, $lm=-91$

solving we get, $l=-6,m=\frac{1}{2}$

Substituting in 2 we get the equations of asymptote to be $x+2y+m=0\Rightarrow 2x+4y+1=0$