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UPSC 2016 Maths Optional Paper 1 Q4d — Step-by-Step Solution

15 marks · Section A

Second-degree equations in three variables · Analytic Geometry · asked 3× in 14 yrs · Read the full method →

Question

Find the locus of the point of intersection of three mutually perpendicular tangent planes to the conicoid $ax^2+by^2+cz^2=1$.

Technique

Tangency condition $p^2=\sum \ell^2/a$; sum it over an orthonormal triad of normals using $\sum(\hat{\mathbf n}_i\cdot\mathbf{OP})^2=|\mathbf{OP}|^2$ and $\sum\ell_i^2=1$.

Solution

Step 1 — Tangency condition for a plane

A plane $\ell x+m y+n z=p$ is tangent to the central conicoid $ax^2+by^2+cz^2=1$ iff

Equivalently, writing the unit normal $\hat{\mathbf n}=(\ell,m,n)/|\mathbf n|$ and perpendicular distance, $(\ast)$ holds for a tangent plane.

Step 2 — Three mutually perpendicular tangent planes through a common point

Let the three tangent planes meet at $P=(x_0,y_0,z_0)$ and have mutually perpendicular unit normals $\hat{\mathbf n}_1,\hat{\mathbf n}_2,\hat{\mathbf n}_3$ (an orthonormal triad). For plane $i$ through $P$, the constant is $p_i=\hat{\mathbf n}_i\cdot\mathbf{OP}$, and the tangency condition $(\ast)$ in unit-normal form reads

where $(\ell_i,m_i,n_i)=\hat{\mathbf n}_i$.

Step 3 — Sum over the orthonormal triad

Add the three equations. On the left,

since the squared projections of a fixed vector onto an orthonormal basis sum to its squared length. On the right, the columns of an orthogonal matrix have unit row-sums of squares:

so

Step 4 — The locus

Equating the two sides and dropping subscripts:

Answer