← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q4b — Step-by-Step Solution

20 marks · Section A

Curve tracing (cartesian and polar) · Calculus · asked 2× in 13 yrs · Read the full method →

Question

Trace the curve $y^2(x^2-1)=2x-1$.

Technique

Solve for $y^2$, find the real-domain (sign analysis), look for symmetry, check asymptotic behaviour at boundaries.

Solution

Step 1 — Solve for $y^2$ and identify the domain.

For real $y$, $y^2\ge 0$. Sign analysis of the right-hand side:

So the curve exists for $x\in(-1,1/2]\cup(1,\infty)$.

Step 2 — Symmetry.

Only $y^2$ appears: curve symmetric about the $x$-axis.

Step 3 — Behaviour at the boundaries.

• As $x\to-1^+$: $(x-1)(x+1)\to 0^-$ (since $x+1\to 0^+,\,x-1<0$); $2x-1\to -3$. So $y^2=(-3)/0^-\to+\infty$. Vertical asymptote $x=-1$. Curve goes to $\pm\infty$.
• At $x=1/2$: $y=0$. Single point on the $x$-axis.
• As $x\to 1^+$: $(x-1)(x+1)\to 0^+$; $2x-1\to 1$. $y^2\to+\infty$. Vertical asymptote $x=1$.
• As $x\to\infty$: $y^2\sim 2x/x^2=2/x\to 0$. Curve approaches the $x$-axis from above and below.

Step 4 — Sketch summary.

• Branch 1: $x\in(-1,1/2]$. Curve has two arcs (above and below $x$-axis). At $x\to -1^+$, both arcs go to $\pm\infty$. They come down monotonically (to be checked, but a single root of $y'=0$ in this interval gives the local extremum) and meet at $(1/2,0)$ on the $x$-axis.

• Branch 2: $x\in(1,\infty)$. Two arcs (above and below). At $x\to 1^+$, both go to $\pm\infty$. Approach $y=0$ as $x\to\infty$. Curve in each branch is monotonic (each $|y|$ decreasing).

The curve has the shape: in the strip $(-1,1/2]$, two infinite arcs joined at a single rightmost point on the $x$-axis. To the right of $x=1$, two infinite arcs going from the asymptote down to the $x$-axis at infinity.

Answer