Question

The curve $y^2(x-2)=x^2(1+x)$ has:

• A. An asymtote parallel to $x$-axis
• B. An asymtote parallel to $y$-axis
• C. Asymtotes parallel to both axes
• D. No asymptote

Answer 1

The correct option is B An asymtote parallel to $y$-axis

To find the asymtote, we need to get one variable in terms of other variable.

By observation, we see that it is very easy to get $y$ in terms of $x$.

So, that's exactly what we will do:

$y^2(x-2)=x^2(1+x)$

$\Rightarrow y^2=\frac{x^2(1+x)}{x-2}$

By definition, asymtote can be found when for a finite value of one co-ordinate, other tends to $\infty$ or $-\infty$.

So, we see when $x=2$, $y=\infty$.

So, $x=2$ is an asymptote which is parallel to $y$-axis.