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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 the other asymptote is
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Verified Answer
The correct answer is:
$2 x+4 y+1=0$
Given hyperbola,
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, $2 x+4 y+1=0$.
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, $2 x+4 y+1=0$.
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