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UPSC 2026 Maths Optional Paper 1 Q4c-ii — Step-by-Step Solution

10 marks · Section A

Cylinder · Analytic Geometry · asked 5× in 14 yrs · Read the full method →

Question

Find the equation of the right circular cylinder whose axis is $x = 2y = -z$, and radius is $4$. Prove that the area of the section of this cylinder by the plane $z = 0$ is $24\pi$.

Technique

A right circular cylinder is the locus of points at a fixed distance from its axis, so impose $|\vec r\times\hat d|=4$ with $\hat d$ the unit vector along the axis, using Lagrange’s identity $|\vec r\times\hat d|^2=|\vec r|^2-(\vec r\cdot\hat d)^2$ to keep the algebra clean. For the section, set $z=0$ and identify the resulting central conic through the invariants $\operatorname{tr}$ and $\det$ of its matrix — the eigenvalues give the semi-axes, hence the area. Confirm independently by the parallel-projection area formula.

Solution

Step 1 — The axis in standard form.

So the axis is the line through the origin $O(0,0,0)$ (all three members vanish there) with direction ratios

Step 2 — Equation of the cylinder.

Let $P(x,y,z)$ be any point of the cylinder and $\vec r=\overrightarrow{OP}=(x,y,z)$. The perpendicular distance from $P$ to the axis is $|\vec r\times\hat d|$, and the cylinder is the locus where this equals the radius $4$:

By Lagrange’s identity, $|\vec r\times\vec d|^2=|\vec r|^2|\vec d|^2-(\vec r\cdot\vec d)^2$, so the condition is

Expand the square:

Substituting into $(2)$:

Check. The point $Q=\dfrac{4}{\sqrt5}(1,-2,0)$ satisfies $Q\cdot\vec d=\dfrac{4}{\sqrt5}(2-2-0)=0$ and $|Q|=\dfrac{4}{\sqrt5}\sqrt5=4$, so it is at distance $4$ from the axis and must lie on the cylinder. Indeed

(The equation is automatically invariant under $\vec r\mapsto\vec r+s\vec d$, since $(\vec r+s\vec d)\times\vec d=\vec r\times\vec d$ — as it must be for a cylinder.)

Step 3 — The section by $z=0$: identify the conic.

Setting $z=0$ in the cylinder’s equation, the section is the plane curve

Write it as $ax^2+2hxy+by^2=144$ with $a=5,\ h=-2,\ b=8$. There are no linear terms, so it is a central conic with centre at the origin (the axis meets $z=0$ at $O$, as it must). Its discriminant is

so $(3)$ is an ellipse (real, since $a>0$ and the right-hand side $144>0$). It is not a circle, because the plane $z=0$ is not perpendicular to the axis.

Step 4 — Semi-axes by the invariants of the conic.

The matrix of the quadratic form is

Both are invariant under rotation of axes, so if $\lambda_1,\lambda_2$ are the eigenvalues of $M$ then in principal axes $(u,v)$ the conic is $\lambda_1u^2+\lambda_2v^2=144$. The characteristic equation is

Both eigenvalues are positive, confirming a real ellipse. Hence

so the semi-axes are

Therefore

Equivalently, in one line: for $\lambda_1u^2+\lambda_2v^2=k$ the area is $\pi\sqrt{\frac{k}{\lambda_1}}\sqrt{\frac{k}{\lambda_2}}=\dfrac{\pi k}{\sqrt{\lambda_1\lambda_2}}=\dfrac{\pi k}{\sqrt{\det M}}=\dfrac{144\pi}{6}=24\pi$.

Where the axes point (a geometric confirmation). For $\lambda=9$: $(M-9I)w=0$ gives $-4w_1-2w_2=0$, i.e. $w\propto(1,-2)$; for $\lambda=4$: $w_1-2w_2=0$, i.e. $w\propto(2,1)$. So the minor axis (half-length $4$) lies along $(1,-2,0)$ — and $(1,-2,0)\cdot(2,1,-2)=2-2+0=0$, so this direction is perpendicular to the cylinder’s axis, exactly as it must be: along it the section coincides with the right circular cross-section, giving half-length equal to the radius $4$. $\checkmark$ The major axis (half-length $6$) lies along $(2,1,0)$. Both endpoints check on $(3)$:

Step 5 — Independent proof of the area by parallel projection.

Lemma. Let $\hat d$ be a unit vector, $\Pi_1$ a plane with unit normal $\hat n_1$ where $\hat n_1\cdot\hat d\ne 0$, and $\Pi_2$ a plane perpendicular to $\hat d$. Projection along $\hat d$ from $\Pi_1$ to $\Pi_2$ multiplies areas by $|\hat n_1\cdot\hat d|$.

Proof. A parallelogram in $\Pi_1$ spanned by $\vec u,\vec v$ maps to the one spanned by $\vec u\,'=\vec u-(\vec u\cdot\hat d)\hat d$, $\vec v\,'=\vec v-(\vec v\cdot\hat d)\hat d$. Both lie in $\Pi_2$, so $\vec u\,'\times\vec v\,'$ is parallel to $\hat d$ and

since subtracting multiples of the third column from the first two leaves a determinant unchanged. Hence

Application. Every generator of the cylinder is parallel to $\hat d=\tfrac13(2,1,-2)$, and $\hat d\cdot\hat k=-\tfrac23\ne 0$, so each generator meets the plane $z=0$ in exactly one point. Projection along $\hat d$ therefore carries the section by $z=0$ bijectively onto the right circular cross-section — a circle of radius $4$, of area $16\pi$. With $\hat n_1=\hat k=(0,0,1)$,

so by the Lemma

The two independent methods agree, and the semi-axes $4$ (the radius, unchanged) and $6=4\div\tfrac23$ (the radius stretched by the obliquity factor) are exactly what the projection picture predicts.

Answer