UPSC 2020 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Asymptotes · Calculus · Read the full method →

Question

Find all the asymptotes of the curve $(2x+3)y=(x-1)^2$ .

Technique

Write $y$ explicitly; vertical asymptote at zero of denominator; oblique asymptote via long division (or $m=\lim y/x$ , $c=\lim(y-mx)$ ).

Solution

Solve for $y$ :

y=\frac{(x-1)^2}{2x+3}=\frac{x^2-2x+1}{2x+3}.

The curve is a rational function (degree of numerator $2$ , denominator $1$ ): expect one vertical asymptote and one oblique asymptote (no horizontal asymptote, since numerator degree $>$ denominator degree).

Step 1 — Vertical asymptote

$y\to\infty$ where the denominator vanishes but the numerator does not. $2x+3=0\Rightarrow x=-\tfrac32$ ; at $x=-\tfrac32$ the numerator is $(-\tfrac32-1)^2=\tfrac{25}{4}\neq0$ . Hence

\boxed{x=-\tfrac32}\ \text{(vertical asymptote).}

Step 2 — Oblique asymptote by polynomial division

Divide $x^2-2x+1$ by $2x+3$ :

\frac{x^2-2x+1}{2x+3}=\frac{x}{2}-\frac{7}{4}+\frac{25/4}{2x+3}.

Check of division: $\left(\tfrac{x}{2}-\tfrac74\right)(2x+3)=x^2+\tfrac{3x}{2}-\tfrac{7x}{2}-\tfrac{21}{4}=x^2-2x-\tfrac{21}{4}$ ; adding the remainder $\tfrac{25}{4}$ gives $x^2-2x+1$ ✓.

As $x\to\pm\infty$ , the term $\dfrac{25/4}{2x+3}\to0$ , so

y\to\frac{x}{2}-\frac{7}{4}.

\boxed{y=\tfrac12 x-\tfrac74,\ \text{i.e. } 4y=2x-7\quad\text{(oblique asymptote).}}

Step 3 — Complete list

Answer

\boxed{\;x=-\frac32\quad\text{and}\quad y=\frac{x}{2}-\frac{7}{4}.\;}