Obtain The Asymptotes of The Curve #
\(\pmb{ y^{3}+x^{2}y+2xy^{2}-y+1=0}\) #
Answer:
The given curve is \begin{align*} \pmb{ y^{3}+x^{2}y+2xy^{2}-y+1=0} \end{align*}
(a) To find the vertical asymptotes:
The coefficent of the highest power of \(\pmb{ y}\) (i.e., \(\pmb{ y^{3}}\) ) is \(\pmb{ 1}\) . Therefore there is no vertical asymptote.
(b) To find the horizontal and oblique asymptotes:
Putting \(\pmb{ x=1}\) and \(\pmb{ y=m}\) in the 3rd degree terms. we get
\begin{align}
\pmb{ \phi_{3}(m)=m^{3}+2m^{2}+m}
\end{align}
and
\begin{align}
&\pmb{ \phi_{3}(m)= 0}\nonumber\\
\implies &\pmb{ m^{\prime}+2m^{2}+m= 0}\nonumber\\
\implies &\pmb{ m(m^{2}+2m+1)= 0}\nonumber\\
\implies &\pmb{ m(m+1)^{2}= 0}\nonumber\\
\implies &\pmb{ m=0,-1,-1}
\end{align}
Again from (1), we get
\begin{align}
&\pmb{ \phi^{\prime}_{3}(m)=3m^{2}+4m+1}\\
&\pmb{ \phi^{\prime\prime}_{3}(m)=6m+4}
\end{align}
Putting \(\pmb{ x=1}\) and \(\pmb{ y=m}\) in the 2nd degree terms. we get
\begin{align}
&\pmb{ \phi_{2}(m)=0}\\
&\pmb{ \phi^{\prime}_{2}(m)=0}\\
\end{align}
Putting \(\pmb{ x=1}\) and \(\pmb{ y=m}\) in the 1st degree terms. we get
\begin{align}
\pmb{ \phi_{1}(m)=-m}
\end{align}
Putting \(\pmb{ x=1}\) and \(\pmb{ y=m}\) in the zero degree terms. we get
\begin{align}
\pmb{ \phi_{0}(m)=1}
\end{align}
For \(\pmb{ m=0}\),
\begin{align}
&\pmb{ c\phi^{\prime}_{3}(0)+\phi_{2}(0)=0}\nonumber\\
\implies&\pmb{ c=0}
\end{align}
Since \( \pmb{ \phi^{\prime}_{3}(0)=1, ~\phi_{2}(0)=0} \) by using (3) and (5)
For \(\pmb{ m=-1}\),
\begin{align}
&\pmb{ \frac{c^{2}}{2}\phi^{\prime\prime}_{3}(-1)+c\phi^{\prime}_{2}(-1)+\phi_{1}(-1)=0}\nonumber\\
\implies&\pmb{ -c^{2}+1=0}\nonumber\\
\implies&\pmb{ c^{2}=1}\nonumber\\
\implies&\pmb{ c=\pm 1}
\end{align}
Since \( \pmb{ \phi^{\prime\prime}_{3}(-1)=-2, ~\phi^{\prime}_{2}(-1)=0, ~\phi_{1}(-1)=1} \) by using (4),(6) and (7)
Hence there are three asymptotes:
- Horizontal Asymptote: \( \pmb{y=0 } \)
- Oblique Asymptote: \( \pmb{y=-x+1 } \)
- Oblique Asymptote: \( \pmb{y=-x-1 } \)
