Question

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

Answer 1

Answer: (B)

Given hyperbola,
$14x^2+38xy+20y^2+x-7y-91=0...(i)$
On factorising $14x^2+38xy+20y^2$, we get
$=(7x+5y)(2x+4y)$
One of the asymptote is $7x+5y-3=0$
Then, let other asymptote is $2x+4y+k=0$
So, on combining
$(7x+5y-3)(2x+4y+k)=0\dots (ii)$
On equating the coefficient of $x$ from Eqs. (i) and (ii), we get
$7k-6=1$
$\Rightarrow k=1$
So, other asymptote is, $2x+4y+1=0$.