Question

Which of the following options is not the asymptote of the curve
$3x^3+2x^{2}y-7x{y}^2+2y^3-14xy+7y^2+4x+5y=0$?

• A. $y=\frac{-1}{2}x-\frac{5}{6}$
• B. $y=x-\frac{7}{6}$
• C. $y=2x+\frac{3}{7}$
• D. $y=3x-\frac{3}{2}$

Answer 1

Answer: (C)

$y=2x+\frac{3}{7}$

${\varphi }_3\left(m\right)=3+2m-7m^2+2m^3$ ............. $(x=1,y=m)$
${\varphi }_2\left(m\right)=-14m+7m^2$
${\varphi }_3^{\prime }\left(m\right)=2-14m+6m^2$
Now, putting ${\varphi }_3\left(m\right)=0$, we have
$3+2m-7m^2+2m^3=0$
$\Rightarrow (1-m)(1+2m)(3-m)=0$
$\Rightarrow m=-\frac{1}{2},1,3$
We know that
$c{\varphi }_n^{\prime }\left(m\right)+{\varphi }_{n-1}\left(m\right)=0$, which in the given case becomes
$c(2-14m+6m^2)+(-14m+7m^2)=0$
$⟹c=\frac{14m-7m^2}{2-14m+6m^2}$
So, when $m=-\frac{1}{2},c=-\frac{5}{6}$
When $m=1,c=-\frac{7}{6}$
When $m=3,c=-\frac{3}{2}$
$\therefore$ Asymptotes are $y=-\frac{1}{2}x-\frac{5}{6}$,
$y=x-\frac{7}{6}$ and $y=3x-\frac{3}{2}$.

Hence, C is not the asymptote of the given curve.