← 2026 Paper 1

UPSC 2026 Maths Optional Paper 1 Q4b — Step-by-Step Solution

15 marks · Section A

Curve tracing (cartesian and polar) · Calculus · asked 3× in 14 yrs · Read the full method →

Question

Trace the curve $y^2(a + x) = x^2(3a - x)$.

Technique

Run the standard tracing checklist on the cubic $F(x,y)\equiv y^2(a+x)-x^2(3a-x)=0$: symmetry, intercepts, region of existence from $y^2=\dfrac{x^2(3a-x)}{a+x}\ge 0$, singular points (the origin is a double point — classify it by the discriminant $H^2-AB$ of the lowest-degree terms), asymptotes (coefficient of the highest power of $y$; then $\phi_3(m)$ for oblique ones), and turning points from $\dfrac{d(y^2)}{dx}=0$. Each item is one bold step so every mark is separately visible to the examiner.

Solution

Throughout take $a>0$ (the constant is understood to be a positive length; the case $a<0$ is the mirror image $x\mapsto -x$, and $a=0$ degenerates the equation to $y^2 = -x^2$, i.e. the single point at the origin). Write

Step 1 — Symmetry.

$y$ occurs only through $y^2$, so $F(x,-y)=F(x,y)$: the curve is symmetric about the $x$-axis.

It is not symmetric about the $y$-axis ($F(-x,y)=ay^2-xy^2-3ax^2-x^3\ne F(x,y)$), nor about the origin, nor about $y=x$. So only the upper half need be traced; the lower half is its reflection.

Step 2 — Solve for $y^2$ and find the region of existence.

Since $x^2\ge 0$, real points require $\dfrac{3a-x}{a+x}\ge 0$ (or $x=0$).

Hence the curve exists only for $-a<x\le 3a$. It lies entirely in the vertical strip between the line $x=-a$ (excluded) and the line $x=3a$ (attained).

Step 3 — Intercepts.

With the $x$-axis: put $y=0$ in $F=0$: $x^2(3a-x)=0\Rightarrow x=0$ (a repeated root) or $x=3a$. So the curve meets the $x$-axis at $O(0,0)$ and at $A(3a,0)$; the double root at $x=0$ already signals that the origin is a singular point.

With the $y$-axis: put $x=0$: $ay^2=0\Rightarrow y=0$. The origin is the only intersection.

There is no other point on the axes.

Step 4 — Singular points; nature of the origin.

A point is singular iff $F=F_x=F_y=0$.

$F_y=0$ gives $y=0$ or $x=-a$; but $x=-a$ carries no point of the curve (Step 2), so $y=0$. Then $F=0$ forces $x^2(x-3a)=0$, i.e. $x=0$ or $x=3a$.

• At $(3a,0)$: $F_x=0-18a^2+27a^2=9a^2\ne 0$, so this is an ordinary (non-singular) point.
• At $(0,0)$: $F_x=0$ and $F_y=0$, so the origin is a double point — the only singular point of the curve.

Classification. At a double point the tangents are given by the lowest-degree (here degree-two) terms of $F$:

Applying the standard criterion: writing the degree-two part as $Ax^2+2Hxy+By^2$ we have $A=-3a$, $H=0$, $B=a$, so

$H^2-AB>0$ $\Rightarrow$ two distinct real tangents $\Rightarrow$ the origin is a node.

(This is consistent with Step 2: real branches of the curve exist on both sides of $x=0$, so the origin cannot be an isolated point; and the two tangents are distinct, so it is not a cusp.) The tangents make angles $\pm 60^\circ$ with the $x$-axis, since $\tan^{-1}\sqrt3 = 60^\circ$.

Step 5 — Tangent at $A(3a,0)$.

At an ordinary point the tangent is $F_x\,(x-3a)+F_y\,(y-0)=0$. Here $F_x=9a^2$, $F_y=0$, so

The tangent at $A(3a,0)$ is vertical: the curve meets the $x$-axis at right angles there and turns back, which is exactly what closes the loop.

Step 6 — Asymptotes.

Parallel to the $y$-axis. Regard $F=0$ as a polynomial in $y$:

Equating the coefficient of the highest power of $y$ to zero: $a+x=0$, i.e.

is a vertical asymptote. (Confirming directly from $(\ast)$: as $x\to -a^{+}$, the numerator $\to a^2\cdot 4a=4a^3>0$ while $a+x\to 0^{+}$, so $y^2\to+\infty$ and $y\to\pm\infty$.)

Parallel to the $x$-axis. The coefficient of the highest power of $x$ (namely $x^3$) is the constant $1\ne 0$, so there is no horizontal asymptote.

Oblique. The degree-three homogeneous part of $F$ is

Put $y=mx$: $\phi_3(m)=F_3(1,m)=1+m^2$. Since $1+m^2>0$ for every real $m$, $\phi_3$ has no real root, so there is no oblique asymptote. (The factorisation $x(x^2+y^2)$ shows the three asymptotic directions: the real one $x=0$ — which produced the vertical asymptote $x=-a$ — and the two imaginary ones $y=\pm ix$ through the circular points at infinity.)

The only asymptote is $x=-a$.

Step 7 — Turning points (extreme values of $|y|$).

Differentiate $(\ast)$ with $N=3ax^2-x^3$, $D=a+x$:

Numerator $=6a^2x+6ax^2-3ax^2-3x^3-3ax^2+x^3=6a^2x-2x^3=2x\,(3a^2-x^2)$. Hence

Setting $\dfrac{dy}{dx}=0$: $x=0$ or $x=\pm\sqrt3\,a$. Of these, $x=-\sqrt3\,a<-a$ is outside the region of existence, and $x=0$ is the node (where $(\dagger)$ is $0/0$ and Step 4 gives the two tangents instead). The genuine turning point is

Sign of $(\dagger)$ (monotonicity), noting $(a+x)^2>0$ and taking the upper branch $y>0$:

So on the upper half there is exactly one maximum, at $\left(\sqrt3\,a,\;a\sqrt{6\sqrt3-9}\right)$, and by symmetry one minimum at $\left(\sqrt3\,a,\;-a\sqrt{6\sqrt3-9}\right)$.

Step 8 — Table of points (take $a=1$; scale by $a$ in general).

(Each entry is $(\ast)$ evaluated directly; e.g. $x=1$: $y^2=\frac{1\cdot 2}{2}=1$, and indeed $y^2(a+x)=1\cdot 2=2=x^2(3a-x)$. $\checkmark$)

Step 9 — The traced curve.

Assembling Steps 1–8:

• The curve lies wholly in the strip $-a<x\le 3a$ and is symmetric about the $x$-axis.
• A loop in $0\le x\le 3a$: starting at the node $O$ along the tangent $y=\sqrt3\,x$ (rising at $60^\circ$), the upper arc climbs to the highest point $\left(\sqrt3\,a,\;a\sqrt{6\sqrt3-9}\right)\approx(1.732a,\,1.180a)$, then falls and meets the $x$-axis perpendicularly at $A(3a,0)$. Its mirror image below the axis returns to $O$ along $y=-\sqrt3\,x$. The two arcs together form a closed loop of width $3a$ and greatest height $\approx 1.18a$, bulging to the right of centre (its widest point is at $x=\sqrt3\,a\approx 1.732a = 0.577\times 3a$, i.e. right of the loop’s mid-abscissa $1.5a$).
• Two infinite branches in $-a<x<0$: from the node $O$, one arc leaves along $y=-\sqrt3\,x$ into the second quadrant, rising monotonically to $+\infty$ as $x\to -a^{+}$; its mirror image leaves along $y=+\sqrt3\,x$ into the third quadrant, falling to $-\infty$. Both hug the vertical asymptote $x=-a$, which they never meet.
• The two branch-pairs cross transversally at the origin — that crossing is the node, with the loop’s tangents and the infinite branches’ tangents being the same pair of lines $y=\pm\sqrt3\,x$.

The whole figure looks like a loop to the right of the origin with a pair of horns opening leftwards to infinity against $x=-a$. (This cubic is the classical trisectrix of Maclaurin.)

Figure specification (what the sketch must carry to earn the marks — these are the mathematical contents, not decoration):

1. Axes with $O$ marked, and the points $A(3a,0)$ and $(-a,0)$ labelled on the $x$-axis.
2. The vertical asymptote $x=-a$ drawn dashed, with both infinite branches approaching it, one to $+\infty$ and one to $-\infty$, and never touching it.
3. The two tangent lines $y=\pm\sqrt3\,x$ at the origin drawn dashed and short, labelled, so the node is unmistakable; the four arcs must visibly leave $O$ along them at $\pm 60^\circ$.
4. The loop closed at $A(3a,0)$ with a visibly vertical tangent there.
5. The turning points $\left(\sqrt3\,a,\;\pm a\sqrt{6\sqrt3-9}\right)$ marked with a dot and labelled; the loop’s maximum height must sit at $x=\sqrt3\,a\approx1.732a$, which is right of the loop’s mid-abscissa $1.5a$ — the loop is visibly skewed towards $A(3a,0)$, not symmetric about $x=1.5a$.
6. Nothing drawn for $x<-a$ or $x>3a$ — the empty regions are part of the answer.

Answer