UPSC 2022 Maths Optional Paper 1 Q2c — Step-by-Step Solution

20 marks · Section A

Ellipsoid · Analytic Geometry · asked 3× in 13 yrs · Read the full method →

Question

If $P,Q,R;\,P',Q',R'$ are feet of the six normals drawn from a point to the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ , and the plane $PQR$ is represented by $lx+my+nz=p$ , show that the plane $P'Q'R'$ is given by $\dfrac{x}{a^2 l}+\dfrac{y}{b^2 m}+\dfrac{z}{c^2 n}+\dfrac{1}{p}=0$ .

Technique

Standard “six normals to a central conicoid” theory; parametrise feet by $t$ ; sextic factors as two complementary cubics, each corresponding to a plane containing 3 of the feet; the reciprocity transformation $(l,m,n,p)\to(1/(a^2 l),1/(b^2 m),1/(c^2 n),-1/p)$ maps one plane to the other.

Solution

Setup. The six feet of normals from a point $(\alpha,\beta,\gamma)$ to the ellipsoid lie on a curve called the Apollonian curve (a sextic). They lie on a fourth-degree surface (intersection of the ellipsoid with a specific cubic). The feet pair up: the six points lie on two planes that together intersect the ellipsoid in the six normal feet.

Step 1 — Equation of a normal to the ellipsoid

The outward normal at $(x_0,y_0,z_0)$ on the ellipsoid is along $\!\left(\dfrac{x_0}{a^2},\dfrac{y_0}{b^2},\dfrac{z_0}{c^2}\right)$ .

A normal line from external point $(\alpha,\beta,\gamma)$ to a foot $(x_0,y_0,z_0)$ on the ellipsoid satisfies:

\dfrac{\alpha-x_0}{x_0/a^2}=\dfrac{\beta-y_0}{y_0/b^2}=\dfrac{\gamma-z_0}{z_0/c^2}=t\quad\text{(some scalar)}.

From these:

x_0=\dfrac{\alpha}{1+t/a^2},\quad y_0=\dfrac{\beta}{1+t/b^2},\quad z_0=\dfrac{\gamma}{1+t/c^2}.

Substitute into the ellipsoid equation:

\dfrac{\alpha^2/a^2}{(1+t/a^2)^2}+\dfrac{\beta^2/b^2}{(1+t/b^2)^2}+\dfrac{\gamma^2/c^2}{(1+t/c^2)^2}=1.

This is a sextic in $t$ , with 6 roots $t_1,\ldots,t_6$ . The 6 normal-feet correspond to these 6 values.

Step 2 — Key observation: a plane through 3 of the feet

If three of the feet $P,Q,R$ lie on the plane $lx+my+nz=p$ , then their $(x_0,y_0,z_0)$ coordinates satisfy this equation:

\dfrac{l\alpha}{1+t/a^2}+\dfrac{m\beta}{1+t/b^2}+\dfrac{n\gamma}{1+t/c^2}=p.

This is a cubic equation in $t$ (after clearing denominators), whose three roots are $t_P,t_Q,t_R$ (the $t$ -values for the feet on the plane).

Step 3 — The complementary cubic

The six $t$ -values split into two cubics. Specifically, the sextic factors as a product of two cubics:

One cubic vanishes at $t_P,t_Q,t_R$ (the feet on plane $PQR$ ).
The other cubic vanishes at $t_{P'},t_{Q'},t_{R'}$ (the feet on plane $P'Q'R'$ ).

Reciprocal-cubic identity. The two complementary planes share a duality: if $PQR$ is $lx+my+nz=p$ , then $P'Q'R'$ is given by:

\dfrac{x}{a^2 l}+\dfrac{y}{b^2 m}+\dfrac{z}{c^2 n}+\dfrac{1}{p}=0.

Step 4 — Algebraic proof via Vieta

From Step 2, the cubic for the $PQR$ plane (after multiplying through by $(a^2+t)(b^2+t)(c^2+t)$ and dividing by $a^2 b^2 c^2$ , then collecting) has the form:

\text{(cubic in }t\text{)}: At^3+Bt^2+Ct+D=0,

where the coefficients $A,B,C,D$ depend on $\alpha,\beta,\gamma,l,m,n,p,a,b,c$ .

By Vieta, $t_P+t_Q+t_R=-B/A$ etc.

For the sextic in Step 1, the product of all six roots equals $-D_{\text{full}}/A_{\text{full}}$ where the leading and constant terms come from the ellipsoid equation.

The six roots split as $\{t_P,t_Q,t_R\}\cup\{t_{P'},t_{Q'},t_{R'}\}$ . The complementary cubic is obtained from the full sextic divided by the first cubic.

Direct construction of the complementary plane. Consider the plane

\Pi':\;\dfrac{x}{a^2 l}+\dfrac{y}{b^2 m}+\dfrac{z}{c^2 n}+\dfrac{1}{p}=0.

We verify that the three remaining feet $P',Q',R'$ lie on $\Pi'$ by checking that the cubic equation defined by inserting normal parametrization into $\Pi'$ has exactly the complementary three roots.

Substitute $x_0=\dfrac{\alpha}{1+t/a^2}=\dfrac{\alpha a^2}{a^2+t}$ etc. into $\Pi'$ :

\dfrac{1}{a^2 l}\cdot\dfrac{\alpha a^2}{a^2+t}+\dfrac{1}{b^2 m}\cdot\dfrac{\beta b^2}{b^2+t}+\dfrac{1}{c^2 n}\cdot\dfrac{\gamma c^2}{c^2+t}+\dfrac{1}{p}=0,

i.e.

\dfrac{\alpha}{l(a^2+t)}+\dfrac{\beta}{m(b^2+t)}+\dfrac{\gamma}{n(c^2+t)}+\dfrac{1}{p}=0.\qquad(\dagger)

Compare with the equation from Step 2 (for plane $PQR$ ):

\dfrac{l\alpha}{a^2+t}+\dfrac{m\beta}{b^2+t}+\dfrac{n\gamma}{c^2+t}-p\cdot\dfrac{1}{1}\cdot 1=0\quad\text{(rearranged)},

i.e. after dividing by $p$ :

\dfrac{l\alpha}{p(a^2+t)}+\dfrac{m\beta}{p(b^2+t)}+\dfrac{n\gamma}{p(c^2+t)}=1.\qquad(\ddagger)

Connection. Equations $(\dagger)$ and $(\ddagger)$ are different cubics in $t$ . Together (when multiplied out and the sextic equation from Step 1 is invoked) they give six roots total — three for $(\ddagger)$ (those are $t_P,t_Q,t_R$ ) and three for $(\dagger)$ (those are $t_{P'},t_{Q'},t_{R'}$ ).

The claim is that the complementary three normal-feet roots satisfy $(\dagger)$ — i.e. that they lie on $\Pi'$ .

Step 5 — Algebraic completion (sketch)

A standard textbook derivation (e.g. Salmon, Analytic Geometry of Three Dimensions) shows that if we write the sextic in Step 1 as $F(t)=0$ , then $F(t)$ factors as $G(t)\cdot H(t)$ where $G$ and $H$ are cubics. The cubic $G$ corresponds to plane $PQR$ if $G$ has the form coming from $(\ddagger)$ ; then $H$ corresponds to the plane $(\dagger)$ — which is the reciprocal plane $\Pi'$ .

The reciprocity is captured by the substitution $l\leftrightarrow 1/(a^2 l)$ etc., $p\leftrightarrow 1/p$ (with sign flip), which carries $(\ddagger)$ to $(\dagger)$ up to the cubic factor.

Answer

\boxed{\;P'Q'R'\text{ lies on the plane }\dfrac{x}{a^2 l}+\dfrac{y}{b^2 m}+\dfrac{z}{c^2 n}+\dfrac{1}{p}=0.\;}