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

10 marks · Section A

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

Question

Show that the equation $2x^2+3y^2-8x+6y-12z+11=0$ represents an elliptic paraboloid. Also find its principal axis and principal planes.

Technique

Complete the square in $x$ and $y$ separately to bring the equation to canonical paraboloid form; identify vertex (translation), axis (line through vertex along symmetry direction), principal planes (planes of symmetry through axis).

Solution

Step 1 — Complete the square in $x$ and $y$ .

$2(x^2-4x)+3(y^2+2y)-12z+11=0.$

Completing: $2(x-2)^2-8+3(y+1)^2-3-12z+11=0\Rightarrow 2(x-2)^2+3(y+1)^2=12z.$

Divide by $12$ :

\boxed{\;\frac{(x-2)^2}{6}+\frac{(y+1)^2}{4}=z.\;}

This is the canonical form $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=Z$ of an elliptic paraboloid with $a^2=6,\,b^2=4$ , vertex translated to $(X,Y,Z)=(0,0,0)$ , i.e., $(x,y,z)=(2,-1,0)$ .

Step 2 — Identify principal axis and planes.

The paraboloid opens upward (positive- $z$ direction), with axis of symmetry the line through the vertex parallel to the $z$ -axis:

\text{Principal axis: }x=2,\,y=-1.

The principal planes are the planes of symmetry. There are two:

The plane $x=2$ (which contains the principal axis and is parallel to the $yz$ -plane).
The plane $y=-1$ (which contains the principal axis and is parallel to the $xz$ -plane).

Answer

\boxed{\;\text{Principal axis: }x=2,\,y=-1;\quad\text{principal planes: }x=2,\;y=-1.\;}