If the equation of one asymptote of the hyperbola $14 x^2+38 x y+20 y^2+x-7 y-91=0$ is $7 x+5 y-3=0$, then…
- $2 x-4 y+1=0$
- $2 x+4 y+1=0$
- $2 x-4 y-1=0$
- $2 x+4 y-1=0$
Solution

On factorising $14 x^2+38 x y+20 y^2$, we get $ =(7 x+5 y)(2 x+4 y) $ One of the asymptote is $7 x+5 y-3=0$ Then, let other asymptote is $2 x+4 y+k=0$ So, on combining

On equating the coefficient of $x$ from Eqs. (i) and (ii), we get $ 7 k-6=1 \Rightarrow k=1 $ So, other asymptote is, $2 x+4 y+1=0$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)