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UPSC 2021 Maths Optional Paper 1 Q2a — Step-by-Step Solution

20 marks · Section A

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

Question

Show that the planes which cut the cone $ax^2+by^2+cz^2=0$ in perpendicular generators touch the cone $\dfrac{x^2}{b+c}+\dfrac{y^2}{c+a}+\dfrac{z^2}{a+b}=0$.

Technique

Use the standard result for “perpendicular generators” (proved in 2022 P1 Q4c) plus the standard tangency condition for a plane through the vertex of a cone.

Solution

Setup. A plane $ux+vy+wz=0$ through the origin cuts the cone $S:\;ax^2+by^2+cz^2=0$ in two lines (generators). Required: the condition for these generators to be perpendicular implies the plane is tangent to the cone $S':\;\dfrac{x^2}{b+c}+\dfrac{y^2}{c+a}+\dfrac{z^2}{a+b}=0$.

Step 1 — Condition for perpendicular generators

From 2022 P1 Q4(c) we derived: the plane $ux+vy+wz=0$ cuts $ax^2+by^2+cz^2=0$ in perpendicular generators iff

Step 2 — Condition for plane $ux+vy+wz=0$ to touch cone $S'$

A plane $ux+vy+wz=0$ through the origin is tangent to cone $S':\;\dfrac{x^2}{b+c}+\dfrac{y^2}{c+a}+\dfrac{z^2}{a+b}=0$ when the plane and cone intersect along a single line (the generator at the tangent point).

For a cone $\dfrac{x^2}{p}+\dfrac{y^2}{q}+\dfrac{z^2}{r}=0$, the plane $\ell x+my+nz=0$ is tangent iff

(Standard result: a plane through the vertex of a cone is tangent iff it satisfies this dual relation, where $p,q,r$ are the cone’s coefficients.)

Apply to our case: $p=b+c$, $q=c+a$, $r=a+b$, and $(\ell,m,n)=(u,v,w)$:

Tangency condition: $(b+c)u^2+(c+a)v^2+(a+b)w^2=0.$

This is identical to $(\star)$!

Step 3 — Conclude

So the plane $ux+vy+wz=0$ cuts cone $S$ in perpendicular generators iff it satisfies $(\star)$ iff it touches cone $S'$.

Answer