← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q2c-ii — Step-by-Step Solution

10 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

The plane $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ meets the coordinate axes in $A,B,C$ respectively. Prove that the equation of the cone generated by the lines drawn from the origin $O$ to meet the circle $ABC$ is

Technique

A circle in 3D = intersection of a sphere and a plane. Use the sphere $x^2+y^2+z^2=ax+by+cz$ which contains $O,A,B,C$ for free. The cone from $O$ is the locus of rays through $O$ meeting both surfaces; eliminate the scale parameter $t$.

Solution

Step 1 — Identify the sphere through $O,A,B,C$.

$A=(a,0,0),\,B=(0,b,0),\,C=(0,0,c)$.

Consider the sphere with equation $x^2+y^2+z^2=ax+by+cz$, equivalently $x^2+y^2+z^2-ax-by-cz=0$. Verify it passes through all four:

• At $O$: $0=0$ ✓.
• At $A$: $a^2-a^2=0$ ✓.
• $B,C$: similarly ✓.

So this sphere passes through $O,A,B,C$. The circle $ABC$ is the intersection of the sphere $x^2+y^2+z^2=ax+by+cz$ with the plane $x/a+y/b+z/c=1$.

Step 2 — Cone from $O$ through the circle $ABC$.

A point $P=(x,y,z)\ne O$ is on the cone iff the ray from $O$ through $P$ meets the circle $ABC$. Parametrise by $t$: $(tx,ty,tz)$ lies on the plane iff $t(x/a+y/b+z/c)=1$, so $t=abc/(bcx+cay+abz)$. This $t$-scaled point also lies on the sphere (since it’s on the circle, on both surfaces). Substitute:

For $t\ne 0$, divide:

Substitute $t=abc/(bcx+cay+abz)$:

Cross-multiply:

Step 3 — Expand the right-hand side.

$(ax)(bcx)+(ax)(cay)+(ax)(abz)+(by)(bcx)+(by)(cay)+(by)(abz)+(cz)(bcx)+(cz)(cay)+(cz)(abz)$ $=abcx^2+a^2cxy+a^2bxz+b^2cxy+abcy^2+ab^2yz+bc^2xz+ac^2yz+abcz^2.$

Group: $=abc(x^2+y^2+z^2)+(a^2c+b^2c)xy+(a^2b+bc^2)xz+(ab^2+ac^2)yz.$ $=abc(x^2+y^2+z^2)+c(a^2+b^2)xy+b(a^2+c^2)xz+a(b^2+c^2)yz.$

Step 4 — Combine with the LHS.

$abc(x^2+y^2+z^2)$ on the left equals the same on the right, so cancellation gives:

Divide by $abc$:

Recognise $(a^2+b^2)/(ab)=a/b+b/a$, etc.:

Answer