Question

The asymptote of the curve $y^3=x^2(x-a)$ is

• A. $x=0$
• B. $y=0$
• C. $y=x-\frac{a}{3}$
• D. $x=y-\frac{a}{3}$

Answer 1

The correct option is C $y=x-\frac{a}{3}$

Given, $y^3=x^2(x-a)$
$\Rightarrow y=x^{\frac{2}{3}}(x-a{)}^{\frac{1}{3}}$
$\underset{x\to \infty }{lim}y=\infty$ and $\underset{x\to -\infty }{lim}y=-\infty$
So, no horizontal asymptote
$y$ is defined $\forall x\in R,$ so no vertical asymptote.
Let $y=mx+c$ be an oblique asymptote
$m=\underset{x\to \pm \infty }{lim}\left(\frac{y}{x}\right)=\underset{x\to \pm \infty }{lim}\frac{(x-a{)}^{\frac{1}{3}}}{x^{\frac{1}{3}}}=1$
$c=\underset{x\to \pm \infty }{lim}\left(x^{\frac{2}{3}}\right(x-a{)}^{\frac{1}{3}}-x)=-\frac{a}{3}$
$\therefore y=x-\frac{a}{3}$ is an oblique asymptote.