← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q4c — Step-by-Step Solution

15 marks · Section A

Cone · Analytic Geometry · asked 14× in 13 yrs · Read the full method →

Question

If the plane $ux+vy+wz=0$ cuts the cone $ax^2+by^2+cz^2=0$ in perpendicular generators, then prove that $(b+c)u^2+(c+a)v^2+(a+b)w^2=0$.

Technique

Parametrise the line of intersection of plane and cone; treat as a quadratic in slope ratio; use Vieta + perpendicularity condition $l_1 l_2+m_1 m_2+n_1 n_2=0$.

Solution

Setup. Both surfaces pass through the origin. The plane $ux+vy+wz=0$ also passes through the origin (no constant term). The intersection of the plane with the cone consists of two lines through the origin — these are the “generators”.

Let the two generators have direction vectors $\vec\ell_1=(l_1,m_1,n_1)$ and $\vec\ell_2=(l_2,m_2,n_2)$. Each direction must:

1. Lie on the cone: $a l^2+b m^2+c n^2=0$.
2. Lie on the plane: $u l+v m+w n=0$.

Step 1 — Parametrise direction on the plane

From the plane equation: $w n=-(u l+v m)$, so $n=-(ul+vm)/w$ (assuming $w\ne 0$).

Substitute into the cone equation:

Multiply by $w^2$:

Step 2 — Treat as quadratic in $l/m$

Set $t=l/m$. The quadratic is

The two roots $t_1=l_1/m_1$ and $t_2=l_2/m_2$ correspond to the two generators.

By Vieta:

• $t_1+t_2=-\dfrac{2cuv}{aw^2+cu^2}$.
• $t_1 t_2=\dfrac{bw^2+cv^2}{aw^2+cu^2}$.

Step 3 — Perpendicularity condition

$\vec\ell_1\perp\vec\ell_2\Leftrightarrow l_1 l_2+m_1 m_2+n_1 n_2=0$.

Dividing by $m_1 m_2$:

i.e. $t_1 t_2+1+\dfrac{n_1 n_2}{m_1 m_2}=0$.

Step 4 — Compute $n_1 n_2/(m_1 m_2)$

From plane: $n=-(ul+vm)/w=-\dfrac{u t+v}{w}\cdot m\;/m$ — let me redo.

$n_i=-(u l_i+v m_i)/w$, so $n_i/m_i=-(u t_i+v)/w$.

Hence $\dfrac{n_1 n_2}{m_1 m_2}=\dfrac{(u t_1+v)(u t_2+v)}{w^2}=\dfrac{u^2 t_1 t_2+uv(t_1+t_2)+v^2}{w^2}$.

Substitute Vieta:

Wait — let me recompute: $u^2 b w^2+a v^2 w^2=w^2(bu^2+av^2)$.

So $\dfrac{n_1 n_2}{m_1 m_2}=\dfrac{w^2(bu^2+av^2)}{w^2(aw^2+cu^2)}=\dfrac{bu^2+av^2}{aw^2+cu^2}$.

(Dividing both numerator and denominator by $w^2$.)

Step 5 — Assemble perpendicularity

i.e.

Multiply by $(aw^2+cu^2)$:

Expand:

Group by $u^2,v^2,w^2$:

• Coefficient of $u^2$: $c+b=b+c$.
• Coefficient of $v^2$: $c+a=c+a$.
• Coefficient of $w^2$: $b+a=a+b$.

Answer