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UPSC 2026 Maths Optional Paper 1 Q3c — Step-by-Step Solution

15 marks · Section A

Cone · Analytic Geometry · asked 15× in 14 yrs · Read the full method →

Question

If $\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}$ represents one of a set of three mutually perpendicular generators of the cone $5yz - 8zx - 3xy = 0$, then find the equations of the other two.

Technique

Two standard facts do all the work. (1) A homogeneous cone $ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$ admits three mutually perpendicular generators iff $a+b+c=0$ — verify it holds here. (2) The other two generators, being perpendicular to $(1,2,3)$ and passing through the vertex $O$, lie in the plane $x+2y+3z=0$; intersecting that plane with the cone gives a homogeneous quadratic in two of the variables, which factorises into the two required directions.

Solution

Step 0 — Read off the coefficients.

Compare $5yz-8zx-3xy=0$ with the general cone with vertex at the origin,

i.e. $f=\tfrac52$, $g=-4$, $h=-\tfrac32$. The surface is a cone with vertex at the origin, and every line through $O$ lying on it is a generator, so a generator is fully specified by its direction ratios.

Step 1 — Verify that $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ is a generator.

The line passes through the origin with direction ratios $(l_1,m_1,n_1)=(1,2,3)$. Substituting $(x,y,z)=(1,2,3)$ into the cone:

So the given line does lie on the cone; it is a generator.

Step 2 — Check the condition for three mutually perpendicular generators.

Standard result. A cone $ax^{2}+by^{2}+cz^{2}+2fyz+2gzx+2hxy=0$ has three mutually perpendicular generators if and only if $a+b+c=0$.

Why (one line, worth writing): let $A=\begin{pmatrix}a&h&g\\h&b&f\\g&f&c\end{pmatrix}$ so the cone is $\mathbf v^{T}A\mathbf v=0$. If $\mathbf u_1,\mathbf u_2,\mathbf u_3$ are mutually perpendicular unit generators they form an orthonormal basis, so $\sum_i \mathbf u_i\mathbf u_i^{T}=I$ and

Each term on the left is $0$ (each $\mathbf u_i$ is on the cone), hence $a+b+c=0$.

Here $a+b+c=0+0+0=0$, so such a triad does exist — the hypothesis of the question is consistent.

Step 3 — Locate the plane containing the other two generators.

The other two generators pass through the vertex $O$ and are perpendicular to the direction $(1,2,3)$. A line through $O$ with direction $(l,m,n)$ is perpendicular to $(1,2,3)$ iff

So both remaining generators lie in the plane through the origin with normal $(1,2,3)$:

Step 4 — Intersect the plane with the cone.

From $\pi$, $\;x=-2y-3z$. Substitute into $5yz-8zx-3xy=0$:

Divide by $6$:

(If $z=0$ this forces $6y^{2}=0$, so $y=0$ and then $x=0$ — only the origin. Hence no generator in $\pi$ has $z=0$, and we may safely normalise $z=1$ below without losing a direction.)

Step 5 — Extract the two directions.

Case 1: $y+z=0$, i.e. $y=-z$. Then $x=-2y-3z=-2(-z)-3z=-z$. Taking $z=-1$:

Case 2: $y+4z=0$, i.e. $y=-4z$. Then $x=-2(-4z)-3z=8z-3z=5z$. Taking $z=1$:

Step 6 — Verify both lie on the cone.

$(1,1,-1)$: $\;5(1)(-1)-8(-1)(1)-3(1)(1)=-5+8-3=0.\ \checkmark$

$(5,-4,1)$: $\;5(-4)(1)-8(1)(5)-3(5)(-4)=-20-40+60=0.\ \checkmark$

Step 7 — Verify mutual perpendicularity of all three.

The three directions are pairwise orthogonal and all three lie on the cone: they are a set of three mutually perpendicular generators.

Remark (why the third check could not have failed). Steps 3–5 force the last two directions to be perpendicular to $(1,2,3)$, but perpendicularity to each other looks like luck. It is not. Take an orthonormal basis $\{\mathbf u_1,\mathbf e,\mathbf f\}$ of $\mathbb R^{3}$ with $\mathbf u_1=\frac{1}{\sqrt{14}}(1,2,3)$, so $\{\mathbf e,\mathbf f\}$ spans $\pi$. Writing the restricted form as $Q|_{\pi}=\alpha X^{2}+2\beta XY+\gamma Y^{2}$, the trace argument of Step 2 gives $Q(\mathbf u_1)+\alpha+\gamma=\operatorname{tr}A=0$, and $Q(\mathbf u_1)=0$ since $\mathbf u_1$ is a generator; hence $\alpha+\gamma=0$. A pair of lines $\alpha X^{2}+2\beta XY+\gamma Y^{2}=0$ is perpendicular exactly when $\alpha+\gamma=0$ — so the two remaining generators are automatically orthogonal. This is the same fact $a+b+c=0$ doing its work a second time.

Step 8 — Write the equations.

Both lines pass through the vertex $(0,0,0)$, so in symmetric form:

Answer

(equivalently $x=y=-z$ and $\frac x5=\frac{y}{-4}=z$; together with $\frac x1=\frac y2=\frac z3$ these are three mutually perpendicular generators of $5yz-8zx-3xy=0$.)