← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q3b — Step-by-Step Solution

15 marks · Section A

Paraboloid (elliptic and hyperbolic) · Analytic Geometry · asked 6× in 13 yrs · Read the full method →

Question

Prove that, in general, three normals can be drawn from a given point to the paraboloid x2+y2=2azx^2+y^2=2az, but if the point lies on the surface 27a(x2+y2)+8(a−z)3=027a(x^2+y^2)+8(a-z)^3=0 then two of the three normals coincide.

Technique

Parametrize the normal by λ\lambda, reduce the “foot on surface” condition to a cubic in λ\lambda (three roots →\to three normals); repeated root   ⟺  \iff discriminant =0=0, which factors into the stated surface.

Solution

Step 1 — Normal line at a point of the paraboloid

Let F(x,y,z)=x2+y2−2azF(x,y,z)=x^2+y^2-2az. The normal direction at a surface point is ∇F=(2x,2y,−2a)∥(x,y,−a)\nabla F=(2x,2y,-2a)\parallel(x,y,-a). Let the foot of a normal be (x,y,z)(x,y,z) and the given external point be (α,β,γ)(\alpha,\beta,\gamma). The line from (α,β,γ)(\alpha,\beta,\gamma) along (x,y,−a)(x,y,-a) to the foot gives, for some scalar λ\lambda,

α−x=−λx,β−y=−λy,γ−z=−λ(−a)=aλ,\alpha-x=-\lambda x,\quad \beta-y=-\lambda y,\quad \gamma-z=-\lambda(-a)=a\lambda,

hence

x=α1−λ,y=β1−λ,z=γ−aλ.x=\frac{\alpha}{1-\lambda},\qquad y=\frac{\beta}{1-\lambda},\qquad z=\gamma-a\lambda.

Step 2 — Impose that the foot lies on the paraboloid

Substitute into x2+y2=2azx^2+y^2=2az:

α2+β2(1−λ)2=2a(γ−aλ).\frac{\alpha^2+\beta^2}{(1-\lambda)^2}=2a(\gamma-a\lambda).

Multiply by (1−λ)2(1-\lambda)^2:

α2+β2=2a(γ−aλ)(1−λ)2.\alpha^2+\beta^2=2a(\gamma-a\lambda)(1-\lambda)^2.

Expand the right-hand side. Writing p2:=α2+β2p^2:=\alpha^2+\beta^2, this rearranges to a cubic in λ\lambda:

2a2λ3−(4a2+2aγ)λ2+(2a2+4aγ)λ+(α2+β2−2aγ)=0.(1)2a^2\lambda^3-(4a^2+2a\gamma)\lambda^2+(2a^2+4a\gamma)\lambda+(\alpha^2+\beta^2-2a\gamma)=0. \tag{1}

Step 3 — In general, three normals

Equation (1) is a genuine cubic in λ\lambda (leading coefficient 2a2≠02a^2\neq0). A real cubic has, in general, three roots λ1,λ2,λ3\lambda_1,\lambda_2,\lambda_3, each producing one normal foot (x,y,z)(x,y,z). Hence in general three normals can be drawn from (α,β,γ)(\alpha,\beta,\gamma) to the paraboloid.

Step 4 — Two normals coincide   ⟺  \iff repeated root   ⟺  \iff discriminant =0=0

Two of the three normals coincide precisely when (1) has a repeated root, i.e. its discriminant vanishes. Computing the discriminant of the cubic (sympy) and factoring:

Δ=−4a3 (α2+β2) [ 8a3−24a2γ+24aγ2−8γ3+27a(α2+β2) ].\Delta=-4a^3\,(\alpha^2+\beta^2)\,\Big[\,8a^3-24a^2\gamma+24a\gamma^2-8\gamma^3+27a(\alpha^2+\beta^2)\,\Big].

Discard the factors a3a^3 (constant, a≠0a\neq0) and α2+β2\alpha^2+\beta^2 (its vanishing means the point is on the axis, a degenerate special case). The remaining factor must vanish:

8a3−24a2γ+24aγ2−8γ3+27a(α2+β2)=0.8a^3-24a^2\gamma+24a\gamma^2-8\gamma^3+27a(\alpha^2+\beta^2)=0.

Recognise 8a3−24a2γ+24aγ2−8γ3=8(a−γ)38a^3-24a^2\gamma+24a\gamma^2-8\gamma^3=8(a-\gamma)^3. Therefore

27a(α2+β2)+8(a−γ)3=0.27a(\alpha^2+\beta^2)+8(a-\gamma)^3=0.

Replacing (α,β,γ)(\alpha,\beta,\gamma) by (x,y,z)(x,y,z) (the given point), the coincidence condition is exactly the printed surface:

Answer

  27a(x2+y2)+8(a−z)3=0.  \boxed{\;27a(x^2+y^2)+8(a-z)^3=0.\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.