Second-degree equations in three variables

At a Glance

• Frequency: 3 sub-parts across 2 of 13 years (2016, 2017)
• Priority tier: T3
• Marks (count): 10 (2), 15 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 2, hard 1
• Section: A | Dominant type: derivation

Why This Chapter Matters

Two distinct question types hide under this atom: the director sphere (finding the locus of intersection of three mutually perpendicular tangent planes to a conicoid) and ruled-surface generation (finding the surface traced by a moving line meeting two guide lines under a planarity constraint). The director sphere has appeared in identical form in both 2016 and 2017 — knowing the four-step orthonormal-triad proof cold guarantees 10–15 marks whenever it appears. Ruled-surface generation is rarer but tests the same parametric-elimination technique used throughout analytic geometry, so it pays for itself by cross-training other archetypes.

Minimum Theory

The central conicoid and its tangent planes. For the conicoid $ax^2+by^2+cz^2=1$, a plane $\ell x+my+nz=p$ is tangent to it if and only if $p^2=\frac{\ell^2}{a}+\frac{m^2}{b}+\frac{n^2}{c}.$ This is the standard tangency condition. Written in terms of the unit normal $\hat{\mathbf n}=(\ell,m,n)$ (so $\ell^2+m^2+n^2=1$), the distance from the origin to the tangent plane is $|p|$ and the condition reads $p^2=\ell^2/a+m^2/b+n^2/c$.

Orthonormal triad identity. If $\{\hat{\mathbf n}_1,\hat{\mathbf n}_2,\hat{\mathbf n}_3\}$ is an orthonormal basis of $\mathbb R^3$ (i.e. mutually perpendicular unit vectors), then for any fixed vector $\mathbf v$: $\sum_{i=1}^{3}(\hat{\mathbf n}_i\cdot\mathbf v)^2=|\mathbf v|^2,\qquad\sum_{i=1}^{3}\ell_i^2=\sum_{i=1}^{3}m_i^2=\sum_{i=1}^{3}n_i^2=1.$ These two identities are the engine of the director-sphere derivation.

Ruled surface generation. A ruled surface is generated by a one-parameter family of straight lines. To find its equation: parametrise the generator by its two meeting-point parameters on the guide curves; impose any external constraint (such as parallel to a plane) to reduce to one free parameter; then eliminate all parameters to get a single algebraic relation in $(x,y,z)$.

Question Archetypes

director-sphere (2 question(s); 2016, 2017)

Recognition Cues

• “Find the locus of the point of intersection of three mutually perpendicular tangent planes” to any central conicoid.
• The conicoid is always $ax^2+by^2+cz^2=1$ (a general central quadric without cross terms).
• The answer is always a sphere centred at the origin.

Solution Template

1. State the tangency condition. A plane $\ell x+my+nz=p$ is tangent to $ax^2+by^2+cz^2=1$ iff $p^2=\ell^2/a+m^2/b+n^2/c$.
2. Work with unit normals. Normalise: $\hat{\mathbf n}_i=(\ell_i,m_i,n_i)$ with $|\hat{\mathbf n}_i|=1$ for $i=1,2,3$, and the three normals are mutually perpendicular.
3. Write the tangency condition for each plane through $P=(x_0,y_0,z_0)$. $p_i=\hat{\mathbf n}_i\cdot\mathbf{OP}$ and $p_i^2=\ell_i^2/a+m_i^2/b+n_i^2/c$.
4. Sum the three conditions. Use (i) $\sum_i p_i^2=|\mathbf{OP}|^2=x_0^2+y_0^2+z_0^2$ and (ii) $\sum_i\ell_i^2=\sum_i m_i^2=\sum_i n_i^2=1$.
5. Conclude. The locus is $x^2+y^2+z^2=1/a+1/b+1/c$.

Worked Example(s)

2016 Paper 1, 2016-P1-Q4d (15 marks)

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

Step 1 — Tangency condition. A plane $\ell x+my+nz=p$ is tangent to the conicoid iff $p^2=\frac{\ell^2}{a}+\frac{m^2}{b}+\frac{n^2}{c}.$

Step 2 — Unit-normal setup. Let the three tangent planes have mutually perpendicular unit normals $\hat{\mathbf n}_i=(\ell_i,m_i,n_i)$, $i=1,2,3$, meeting at $P=(x_0,y_0,z_0)$. For each plane through $P$: $p_i=\ell_i x_0+m_i y_0+n_i z_0$.

Step 3 — Three tangency equations. Since $|\hat{\mathbf n}_i|=1$: $p_i^2=\frac{\ell_i^2}{a}+\frac{m_i^2}{b}+\frac{n_i^2}{c},\qquad i=1,2,3.$

Step 4 — Sum. Add all three equations. Left side: $\sum_i p_i^2=\sum_i(\hat{\mathbf n}_i\cdot\mathbf{OP})^2=|\mathbf{OP}|^2=x_0^2+y_0^2+z_0^2$ (Parseval identity for an orthonormal basis). Right side: $\sum_i\frac{\ell_i^2}{a}+\frac{m_i^2}{b}+\frac{n_i^2}{c}=\frac{\sum\ell_i^2}{a}+\frac{\sum m_i^2}{b}+\frac{\sum n_i^2}{c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c},$ since $\sum_i\ell_i^2=\sum_i m_i^2=\sum_i n_i^2=1$ for an orthonormal basis.

Step 5 — Locus. Equating: $\boxed{\;x^2+y^2+z^2=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.\;}$

This is the director sphere of the conicoid, centred at the origin with radius $\sqrt{1/a+1/b+1/c}$. $\blacksquare$

2017 Paper 1, 2017-P1-Q3d (10 marks)

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

This is the identical question to 2016-P1-Q4d. The derivation proceeds through the same five steps; the answer is:

$\boxed{\;x^2+y^2+z^2=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.\;}$

Cross-check. For the unit sphere $a=b=c=1$: the formula gives radius $\sqrt{3}$, matching the known circumscribing sphere of a unit-sphere’s mutually perpendicular tangent configuration. $\checkmark$

For an ellipsoid with semi-axes $A,B,C$ (so $a=1/A^2$, etc.), the director sphere has radius $\sqrt{A^2+B^2+C^2}$ — the familiar sum-of-squares formula.

Common Traps

• Use unit normals. The identity $\sum_i p_i^2=|\mathbf{OP}|^2$ holds only when the normals are unit vectors. Without normalisation, each $p_i$ is not the perpendicular distance from the origin, and the completeness sum fails.
• The key identity is $\sum_i(\hat{\mathbf n}_i\cdot\mathbf v)^2=|\mathbf v|^2$, not $\sum_i(\hat{\mathbf n}_i\cdot\mathbf v)=|\mathbf v|$. The squared projection formula (Pythagoras / Parseval) is what collapses three equations into one.
• Radius formula. For the conicoid in the form $ax^2+by^2+cz^2=1$, the radius is $\sqrt{1/a+1/b+1/c}$. If instead the conicoid is written $x^2/A^2+y^2/B^2+z^2/C^2=1$, identify $a=1/A^2$ and the radius becomes $\sqrt{A^2+B^2+C^2}$. Keep the notational convention straight.
• The director sphere exists as a real sphere only when $1/a+1/b+1/c>0$, i.e., at most one of $a,b,c$ is negative.
• The converse (“if three planes meet on this sphere, they can be mutually perpendicular tangent planes”) is not asked; prove only the forward direction.

ruled-surface-generation (1 question(s); 2016)

Recognition Cues

• “Find the surface generated by a line which meets [two given lines] and is parallel to [a plane].”
• You are given two guide lines $L_1, L_2$ and a planarity/direction constraint on the generator.
• The answer is a quadric surface (degree-2).

Solution Template

1. Parametrise each guide line. Express a general point on $L_1$ as $(x_1(t),y_1(t),z_1(t))$ and on $L_2$ as $(x_2(s),y_2(s),z_2(s))$.
2. Write the direction of the generator. $\mathbf d = (x_2(s)-x_1(t),\,y_2(s)-y_1(t),\,z_2(s)-z_1(t))$.
3. Apply the constraint. If the generator is parallel to a plane with normal $\mathbf n$, then $\mathbf d\cdot\mathbf n=0$. Solve for one parameter in terms of the other.
4. Parametrise the generator. Use the two meeting points and one free parameter $u$ along the line: $(X,Y,Z)=(\text{point on }L_1)+u\cdot\mathbf d$.
5. Eliminate all parameters ($t,s,u$) from the three coordinate equations. The result is the surface equation.
6. Verify. Confirm a specific point on a generator satisfies the surface equation.

Worked Example

2016 Paper 1, 2016-P1-Q4a (10 marks)

Find the surface generated by a line which intersects the lines $y=a,\;z=a$ and $x+3z=a,\;y+z=a$ and is parallel to the plane $x+y=0$.

Step 1 — Parametrise the guide lines.

$L_1$: $y=a$, $z=a$. A general point on $L_1$: $(t,\,a,\,a)$.

$L_2$: $x+3z=a$, $y+z=a$. Put $z=s$: then $y=a-s$, $x=a-3s$. Point on $L_2$: $(a-3s,\,a-s,\,s)$.

Step 2 — Generator direction.

$\mathbf d=(a-3s-t,\,-s,\,s-a).$

Step 3 — Apply the planarity constraint. The plane $x+y=0$ has normal $\mathbf n=(1,1,0)$. Parallel to the plane means $\mathbf d\cdot\mathbf n=0$: $(a-3s-t)+(-s)=0\;\Longrightarrow\;t=a-4s.$

With this substitution: $a-3s-t=s$, so $\mathbf d=(s,-s,s-a)$ and the $L_1$ end becomes $(a-4s,a,a)$.

Step 4 — Parametrise the generator.

$(X,Y,Z)=(a-4s+su,\;a-su,\;a+(s-a)u).$

Step 5 — Eliminate $s$ and $u$. Elimination (via resultants or substitution) removes both $s$ and $u$, giving the surface:

$\boxed{\;y^2+xy+yz+zx-2a(x+z)=0.\;}$

Verification. Substituting the parametrisation into $xy+xz+y^2+yz-2ax-2az$ gives $0$ identically in $s,u$. $\checkmark$

Common Traps

• Both parameters must be eliminated. After applying the planarity constraint, you reduce from two free meeting-point parameters to one (say $s$). But the generator itself has another parameter ($u$ along the line). Eliminating only $s$ leaves an identity; eliminating only $u$ gives a two-parameter family, not a surface. You must eliminate both $s$ and $u$.
• Parallelism condition. The condition is $\mathbf d\cdot\mathbf n=0$, where $\mathbf n$ is the plane’s normal vector — not the plane’s equation evaluated at the point. For $x+y=0$, $\mathbf n=(1,1,0)$.
• Spurious factors. Resultant elimination often produces extra factors (like $a$ or $a-Y$). The surface is the degree-2 factor; discard factors that are constants or linear in just one variable.
• The generated surface is always a quadric when the two guide lines and constraint are linear — don’t worry if the algebra looks messy; the final answer simplifies.

Marks-Aware Writing

10-mark questions (director sphere or ruled surface): Write out each step cleanly. For the director sphere, the tangency condition + two orthonormality identities + “sum and equate” is the entire proof — four logical lines with one display equation each. For ruled surface, the parametrisation + constraint-reduction + elimination is the expected structure; a final verification earns the last mark.

15-mark questions (director sphere extended or harder ruled surface): Same as above but with additional detail. For the director sphere at 15 marks, write the tangency condition derivation (not just state it), explain why $\sum_i p_i^2=|\mathbf{OP}|^2$ (cite Pythagoras/completeness), and verify the formula at $a=b=c$ as a special case. For ruled surfaces, show each intermediate step of the elimination and verify by substitution.

Practice Set

• 2024-P1-Q3c (15 m) — — Hint: find the vertex of a cone by setting $\nabla F=0$.
• 2014-P1-Q4a-ii (10 m) — — Hint: complete the square in each variable; cone iff constant vanishes.
• 2018-P1-Q3c (13 m) — — Hint: related conicoid tangent plane problem.
• 2018-P1-Q4c (13 m) — — Hint: cone-from-vertex-curve; vertex $(0,0,1)$, guiding conic $z=0$.
• 2017-P1-Q1d (10 m) — — Hint: related conicoid / second-degree surface problem.
• 2016-P1-Q4b (10 m) — — Hint: perpendicular generators of the cone $3yz-2zx-2xy=0$.