← 2022 Paper 1

UPSC 2022 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 $y^2 x^2=x^2-a^2$, where $a$ is a real constant.

Technique

Standard curve tracing: symmetry, domain, intercepts, asymptotes (horizontal and vertical), monotonicity. The implicit equation $y^2 x^2=x^2-a^2$ rearranges to $y^2=1-a^2/x^2$, exposing the asymptote $y^2=1$ and the cutoff $|x|=|a|$.

Solution

Setup. Rewrite: $y^2=\dfrac{x^2-a^2}{x^2}=1-\dfrac{a^2}{x^2}$.

So $y=\pm\sqrt{1-a^2/x^2}$.

Step 1 — Symmetry

Equation invariant under $x\to-x$ and $y\to-y$ (only even powers). So the curve is symmetric about both axes (symmetric across origin too).

Step 2 — Domain (real values)

Require $y^2\ge 0\Rightarrow 1-a^2/x^2\ge 0\Rightarrow x^2\ge a^2\Rightarrow|x|\ge|a|$.

So the curve exists only for $|x|\ge|a|$ — i.e., $x\le -|a|$ or $x\ge|a|$.

Step 3 — Asymptotic behaviour

As $x\to\pm\infty$: $y^2\to 1-0=1$, so $y\to\pm 1$.

So $y=\pm 1$ are horizontal asymptotes.

As $x\to|a|^+$: $y^2\to 0$, $y\to 0$. So the curve passes through $(\pm a,0)$.

Similarly $(\pm a,0)$ are the only points on the $x$-axis (apart from where the curve meets it).

Step 4 — Vertical tangents?

$y=\sqrt{1-a^2/x^2}$. $dy/dx=\dfrac{1}{2y}\cdot\dfrac{2a^2}{x^3}=\dfrac{a^2}{y x^3}$.

At $x=a^+$, $y\to 0^+$, so $dy/dx\to\infty$. Vertical tangents at $(\pm a,0)$.

By symmetry, also at $(-a,0)$ (vertical tangent).

Step 5 — Y-axis behaviour

The curve has no points with $|x|<|a|$, so doesn’t touch the $y$-axis (unless $a=0$, degenerate).

Step 6 — Convexity / shape

For $x>a$: $y^2=1-a^2/x^2$ is increasing in $x$ (since $a^2/x^2$ decreases). So $y$ (taking positive branch) increases from 0 at $x=a$ to 1 as $x\to\infty$.

By symmetry, the curve in the first quadrant rises from $(a,0)$ to the asymptote $y=1$.

In the second quadrant ($x<-a$, $y>0$): symmetric image — rises from $(-a,0)$ to $y=1$.

Lower halves ($y<0$): reflections about $x$-axis.

Step 7 — Verbal description of curve

• Exists in two disjoint regions: $x\ge|a|$ and $x\le -|a|$.
• Symmetric about both axes.
• Passes through $(\pm a,0)$ with vertical tangent.
• Approaches horizontal asymptotes $y=\pm 1$ as $|x|\to\infty$.
• Looks like two “trumpet-shaped” pieces, each with two branches (upper and lower) converging to $y=\pm 1$.

Step 8 — Sketch description

       y
       ↑
   y=1 ─────  ─────  (asymptote)
        /        \
       /          \
  ────●──────────●────→ x
       \          /
        \        /
   y=-1─────  ─────  (asymptote)
      x=-a      x=a

Two pieces; each looks like a “trumpet” opening from a cusp/point at $x=\pm a$ widening to the horizontal asymptotes.

Answer