UPSC 2017 Maths Optional Paper 1 Q3d — Step-by-Step Solution

10 marks · Section A

Second-degree equations in three variables · Analytic Geometry · asked 3× in 13 yrs · Read the full method →

Question

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

Technique

Tangency condition $p^2=\sum l^2/a$ ; sum over an orthonormal triad of normals using $\sum l_i^2=1$ and $\sum p_i^2=|OP|^2$ .

Solution

Step 1 — Tangent-plane condition for the conicoid

For the central conicoid $ax^2+by^2+cz^2=1$ , a plane $lx+my+nz=p$ is tangent iff

p^2=\frac{l^2}{a}+\frac{m^2}{b}+\frac{n^2}{c}.

(The standard tangency condition; it follows from substituting the plane into the conicoid and demanding a repeated contact.) Write the unit-normal form so $l^2+m^2+n^2=1$ for each plane; then for a tangent plane at distance $p$ from the origin,

p^2=\frac{l^2}{a}+\frac{m^2}{b}+\frac{n^2}{c}.

Step 2 — Three mutually perpendicular tangent planes

Let the three planes have unit normals $\mathbf n_i=(l_i,m_i,n_i)$ , $i=1,2,3$ , mutually perpendicular, meeting at the point $P=(x_0,y_0,z_0)$ . For each plane through $P$ ,

p_i = l_ix_0+m_iy_0+n_iz_0,\qquad p_i^2=\frac{l_i^2}{a}+\frac{m_i^2}{b}+\frac{n_i^2}{c}.

Step 3 — Sum over the orthonormal triad

Because $\{\mathbf n_1,\mathbf n_2,\mathbf n_3\}$ is an orthonormal basis of $\mathbb R^3$ :

\sum_{i=1}^3 l_i^2=\sum_i m_i^2=\sum_i n_i^2=1,\qquad \sum_i p_i^2=\sum_i(l_ix_0+m_iy_0+n_iz_0)^2=x_0^2+y_0^2+z_0^2

(the last because the triad is orthonormal, so the sum of squared components of the fixed vector $(x_0,y_0,z_0)$ over an orthonormal basis equals its squared length). Summing the tangency relations,

\sum_i p_i^2=\frac{\sum_i l_i^2}{a}+\frac{\sum_i m_i^2}{b}+\frac{\sum_i n_i^2}{c}=\frac1a+\frac1b+\frac1c.

Step 4 — The locus

Equating the two expressions for $\sum p_i^2$ :

x_0^2+y_0^2+z_0^2=\frac1a+\frac1b+\frac1c.

Hence the locus (the director sphere) is

Answer

\boxed{\;x^2+y^2+z^2=\frac1a+\frac1b+\frac1c.\;}