← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q4a — Step-by-Step Solution

15 marks · Section A

Reduction of Second-Degree Equation to Canonical Form · Analytic Geometry · Read the full method →

Question

Reduce the following equation to the standard form and hence determine the nature of the conicoid: $x^2+y^2+z^2-yz-zx-xy-3x-6y-9z+21=0$.

Technique

Diagonalize the quadratic-form matrix; rotate to principal axes; the zero eigenvalue removes one square term, forcing a paraboloid; complete squares and translate to read the standard form $Y'^2+Z'^2=4\sqrt3X'$.

Solution

Step 1 — Matrix of the quadratic part

The quadratic terms $x^2+y^2+z^2-yz-zx-xy$ have symmetric matrix

Step 2 — Eigenvalues of $M$

The characteristic equation gives eigenvalues

A zero eigenvalue (with eigenvector $(1,1,1)$) signals that there is no $X^2$ term along that axis — the surface will be a paraboloid (or degenerate), not a central conicoid. Orthonormal eigenvectors:

Step 3 — Rotate to principal axes

Let $(X,Y,Z)$ be coordinates along $\mathbf e_0,\mathbf e_1,\mathbf e_2$, i.e. $(x,y,z)=X\mathbf e_0+Y\mathbf e_1+Z\mathbf e_2$. Substituting (computer-algebra confirmed) turns the equation into

There is no $X^2$ term (the $\lambda=0$ direction), and the surviving linear term in $X$ is $-6\sqrt3\,X$.

Step 4 — Complete the square in $Y,Z$ and absorb the $X$-shift

The constants combine: $21-\tfrac34-\tfrac94=18$. Writing $Y'=Y+\tfrac{\sqrt2}{2},\ Z'=Z+\tfrac{\sqrt6}{2}$:

with $X'=X-\sqrt3$. Multiply by $\tfrac23$:

Answer