Ellipsoid

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2019, 2020, 2022)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1), 20 (1)
• Average solve time: ~14 min
• Difficulty mix: hard 2, medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

The ellipsoid delivers the widest marks spread of any T3 atom in analytic geometry: one question each at 10, 15, and 20 marks, with every archetype appearing exactly once. The two hard-rated questions (2019 and 2022) require genuine technique — normal-chord parametrisation and the classical six-normals sextic — while the 2020 tangent-plane question is a clean 10-mark computation using the pencil-of-planes method. Since the 2022 question is worth 20 marks and is genuinely difficult, prepare the tangent-plane and normal-chord archetypes for exam security, and learn the six-normals result as a statement-level theorem with the key substitution derivation.

Minimum Theory

Standard ellipsoid. $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ with semi-axes $a\geq b\geq c>0$. The outward unit normal at $(x_0,y_0,z_0)$ is proportional to $\nabla F=\bigl(x_0/a^2,\;y_0/b^2,\;z_0/c^2\bigr)$.

Tangent plane and tangency condition. The tangent plane at $(x_0,y_0,z_0)$ is $\dfrac{xx_0}{a^2}+\dfrac{yy_0}{b^2}+\dfrac{zz_0}{c^2}=1$. A plane $\ell x+my+nz=p$ is tangent to the ellipsoid iff $a^2\ell^2+b^2 m^2+c^2 n^2=p^2.$ Pencil of planes through a line. The line $\pi_1=0=\pi_2$ lies in every plane of the form $\pi_1+\lambda\pi_2=0$. Tangency of one such plane is a quadratic in $\lambda$, giving (generically) two tangent planes through the line.

Normal chord. At $P=(x_1,y_1,z_1)$ on the ellipsoid, the normal direction is $\mathbf n=(x_1/a^2,y_1/b^2,z_1/c^2)$. Write $S=\dfrac{x_1^2}{a^4}+\dfrac{y_1^2}{b^4}+\dfrac{z_1^2}{c^4}=|\mathbf n|^2$ and $S'=\dfrac{x_1^2}{a^6}+\dfrac{y_1^2}{b^6}+\dfrac{z_1^2}{c^6}$. Then the length of the normal chord is $2S^{3/2}/S'$, and the distance from $P$ to where the normal meets the $xy$-plane is $c^2\sqrt{S}$.

Six normals from an external point. From a general external point $(\alpha,\beta,\gamma)$, exactly six normals can be drawn to the ellipsoid. The feet of the six normals split into two groups of three, each group lying in a plane. If the plane through the first group is $\ell x+my+nz=p$, the plane through the other group is $\dfrac{x}{a^2\ell}+\dfrac{y}{b^2 m}+\dfrac{z}{c^2 n}+\dfrac{1}{p}=0$.

Question Archetypes

tangent-plane-through-line (1 question(s); 2020)

Recognition Cues

• “Find the (equations of the) tangent plane(s) to the ellipsoid … which pass through the line …”
• The line is given as the intersection of two planes: $\pi_1=0=\pi_2$.
• There are (generically) two tangent planes; the answer is two explicit plane equations.

Solution Template

1. Write the pencil. Any plane through the line $\pi_1=0=\pi_2$ has the form $\pi_1+\lambda\pi_2=0$. Read off the coefficients $\ell,m,n,p$ as functions of $\lambda$.
2. Put the ellipsoid in standard form $x^2/a^2+y^2/b^2+z^2/c^2=1$. Read off $a^2,b^2,c^2$.
3. Apply the tangency condition $a^2\ell^2+b^2m^2+c^2n^2=p^2$.
4. Solve for $\lambda$. You get a quadratic (generically) with two roots $\lambda_1,\lambda_2$.
5. Write the two tangent planes by substituting $\lambda_1$ and $\lambda_2$ into the pencil equation.
6. Verify each plane contains the line and is tangent to the ellipsoid.

Worked Example

2020 Paper 1, 2020-P1-Q1e (10 marks)

Find the equations of the tangent plane to the ellipsoid $2x^2+6y^2+3z^2=27$ which passes through the line $x-y-z=0=x-y+2z-9$.

Step 1 — Pencil of planes. Any plane through the line is $(x-y-z)+\lambda(x-y+2z-9)=0,$ giving $\ell=1+\lambda$, $m=-(1+\lambda)$, $n=2\lambda-1$, $p=9\lambda$.

Step 2 — Standard form. Divide $2x^2+6y^2+3z^2=27$ by 27: $x^2/(27/2)+y^2/(9/2)+z^2/9=1$, so $a^2=27/2$, $b^2=9/2$, $c^2=9$.

Step 3 — Tangency condition. $a^2\ell^2+b^2m^2+c^2n^2=p^2$: $\frac{27}{2}(1+\lambda)^2+\frac{9}{2}(1+\lambda)^2+9(2\lambda-1)^2=(9\lambda)^2.$ The first two terms: $\bigl(\tfrac{27}{2}+\tfrac{9}{2}\bigr)(1+\lambda)^2=18(1+\lambda)^2$. So: $18(1+\lambda)^2+9(2\lambda-1)^2=81\lambda^2.$ Divide by 9: $2(1+\lambda)^2+(2\lambda-1)^2=9\lambda^2$. Expand: $6\lambda^2+3=9\lambda^2$, giving $3\lambda^2=3$.

Step 4 — Solve. $\lambda=\pm1$.

Step 5 — Two tangent planes.

$\lambda=1$: $\ell=2$, $m=-2$, $n=1$, $p=9$: $2x-2y+z=9.$

$\lambda=-1$: $\ell=0$, $m=0$, $n=-3$, $p=-9$: $z=3.$

$\boxed{\;2x-2y+z=9\quad\text{and}\quad z=3.\;}$

Verification. $z=3$: tangency $c^2\cdot 9=81=p^2$ ✓. Substituting $z=3$: $2x^2+6y^2=0\Rightarrow x=y=0$, single contact point $(0,0,3)$ ✓. Plane $2x-2y+z=9$: tangency $\tfrac{27}{2}(4)+\tfrac{9}{2}(4)+9(1)=54+18+9=81$ ✓.

Common Traps

• Read off $a^2, b^2, c^2$ carefully. For $2x^2+6y^2+3z^2=27$, divide through to get the standard form; $a^2=27/2$, $b^2=27/6=9/2$, $c^2=27/3=9$. Arithmetic errors here propagate through the whole calculation.
• Keep $p=9\lambda$ on the correct side. The pencil equation is $\ell x+my+nz=9\lambda$, so $p=9\lambda$ (positive). If you move the constant to the left you get $p=-9\lambda$; either sign convention works as long as the tangency condition $a^2\ell^2+\cdots=p^2$ uses the same $p$.
• $\lambda=-1$ gives a degenerate-looking plane $z=3$ (the $x$ and $y$ coefficients vanish). Don’t discard it — it is a perfectly valid tangent plane.
• Always verify: (a) each plane contains the line, and (b) each satisfies the tangency condition.

normal-chord (1 question(s); 2019)

Recognition Cues

• “Find the length of the normal chord through $P$” — normal chord means the segment of the normal line between its two intersection points with the ellipsoid.
• “Prove that if [length condition], then $P$ lies on [a cone or surface].”
• The condition often involves $PG_k$, the distance from $P$ to where the normal meets a coordinate plane.

Solution Template

1. Set up the normal parametrisation. At $P=(x_1,y_1,z_1)$ on the ellipsoid, the normal is $(x_1+tx_1/a^2,\;y_1+ty_1/b^2,\;z_1+tz_1/c^2)$, $t\in\mathbb R$.
2. Define $S$ and $S'$. $S=x_1^2/a^4+y_1^2/b^4+z_1^2/c^4$; $S'=x_1^2/a^6+y_1^2/b^6+z_1^2/c^6$.
3. Find the second intersection $P'$. Substitute into the ellipsoid; one root is $t=0$ (at $P$); the other is $t=-2S/S'$.
4. Length $PP'$. $|PP'|=|t|\cdot|\mathbf n|=\dfrac{2S}{S'}\cdot\sqrt{S}=\dfrac{2S^{3/2}}{S'}$.
5. Find $PG_3$. Set the $z$-coordinate of the normal to zero: $t=-c^2$. Then $PG_3=c^2\sqrt{S}$.
6. Impose the condition. $PP'=4PG_3$ gives $S=2c^2 S'$. Expand and rearrange to get the locus.

Worked Example

2019 Paper 1, 2019-P1-Q4b (15 marks)

Find the length of the normal chord through $P$ of the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ and prove that if it equals $4PG_3$, where $G_3$ is where the normal meets the $xy$-plane, then $P$ lies on $\dfrac{x^2}{a^6}(2c^2-a^2)+\dfrac{y^2}{b^6}(2c^2-b^2)+\dfrac{z^2}{c^4}=0$.

Let $P=(x_1,y_1,z_1)$ on the ellipsoid. Normal direction $\mathbf n=(x_1/a^2,y_1/b^2,z_1/c^2)$.

Step 1 — Parametrize normal. Points on the normal: $\bigl(x_1+\tfrac{x_1}{a^2}t,\;y_1+\tfrac{y_1}{b^2}t,\;z_1+\tfrac{z_1}{c^2}t\bigr)$.

Step 2 — Define $S, S'$. $S=\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}+\frac{z_1^2}{c^4}=|\mathbf n|^2,\qquad S'=\frac{x_1^2}{a^6}+\frac{y_1^2}{b^6}+\frac{z_1^2}{c^6}.$

Step 3 — Second intersection. Substituting into the ellipsoid and using $P$ is on it ($t=0$ is one root), the other root is $t_2=-2S/S'$.

Step 4 — Chord length. $PP'=|t_2|\cdot|\mathbf n|=\frac{2S}{S'}\cdot\sqrt{S}=\frac{2S^{3/2}}{S'}.$

$\boxed{\;PP'=\frac{2S^{3/2}}{S'}=\frac{2\bigl(x_1^2/a^4+y_1^2/b^4+z_1^2/c^4\bigr)^{3/2}}{x_1^2/a^6+y_1^2/b^6+z_1^2/c^6}.\;}$

Step 5 — $PG_3$. Setting $z_1+\tfrac{z_1}{c^2}t=0$ gives $t=-c^2$, so $PG_3=c^2\sqrt{S}$.

Step 6 — Impose $PP'=4PG_3$. $\frac{2S^{3/2}}{S'}=4c^2\sqrt{S}\;\Longrightarrow\;\frac{2S}{S'}=4c^2\;\Longrightarrow\; S=2c^2 S'.$ Expanding: $\frac{x_1^2}{a^4}+\frac{y_1^2}{b^4}+\frac{z_1^2}{c^4}=2c^2\!\left(\frac{x_1^2}{a^6}+\frac{y_1^2}{b^6}+\frac{z_1^2}{c^6}\right).$ Collect: $x_1^2\!\left(\frac{1}{a^4}-\frac{2c^2}{a^6}\right)+y_1^2\!\left(\frac{1}{b^4}-\frac{2c^2}{b^6}\right)+z_1^2\!\left(\frac{1}{c^4}-\frac{2c^2}{c^6}\right)=0.$ $\frac{x_1^2}{a^6}(a^2-2c^2)+\frac{y_1^2}{b^6}(b^2-2c^2)-\frac{z_1^2}{c^4}=0.$ Multiply by $-1$ (and drop subscripts): $\boxed{\;\frac{x^2}{a^6}(2c^2-a^2)+\frac{y^2}{b^6}(2c^2-b^2)+\frac{z^2}{c^4}=0.\;}\qquad\blacksquare$ This is homogeneous of degree 2 — a cone with vertex at the origin. $\checkmark$

Common Traps

• Do not normalize $\mathbf n$ prematurely. Both $PP'$ and $PG_3$ carry the same factor $|\mathbf n|=\sqrt{S}$, which cancels in the ratio. Normalising first forces you to carry $1/\sqrt{S}$ everywhere and complicates the algebra.
• $k_{G_3}=-c^2$ is independent of $x_1, y_1$. It comes purely from setting the $z$-coordinate to zero: $z_1(1+t/c^2)=0\Rightarrow t=-c^2$. Many students forget that $G_3$ is on the $xy$-plane (not some other plane).
• Sign in the final locus. The computation gives $(a^2-2c^2)/a^6$ but the stated form has $(2c^2-a^2)/a^6$. Both signs describe the same surface; the question’s preferred form uses $+$ for the $z^2$ term, so flip the sign of the whole equation if needed.
• The final locus is a cone (no constant term, homogeneous degree 2) — this is a sanity check.

feet-of-normals (1 question(s); 2022)

Recognition Cues

• “From a point, six normals are drawn to the ellipsoid; the feet $P,Q,R,P',Q',R'$ split into two planes — if $PQR$ is $\ell x+my+nz=p$, find $P'Q'R'$.”
• “Show that the plane of three of the six normal feet is given by the reciprocal transformation.”

Solution Template

1. Parametrise feet by $t$. A foot $(x_0,y_0,z_0)$ satisfies $x_0=\alpha a^2/(a^2+t)$, $y_0=\beta b^2/(b^2+t)$, $z_0=\gamma c^2/(c^2+t)$, where $t$ is a root of the sextic obtained by substituting into the ellipsoid.
2. Plane $PQR$ gives a cubic. Substituting the parametrised feet into $\ell x+my+nz=p$ gives $\frac{\ell\alpha}{a^2+t}+\frac{m\beta}{b^2+t}+\frac{n\gamma}{c^2+t}=p,$ a cubic in $t$ with roots $t_P,t_Q,t_R$.
3. Reciprocal cubic. The full sextic factors as a product of two cubics; the cubic whose roots are $t_{P'},t_{Q'},t_{R'}$ is obtained by the substitution $\ell\to 1/(a^2\ell)$, $m\to 1/(b^2m)$, $n\to 1/(c^2n)$, $p\to -1/p$ in the plane equation.
4. Write the reciprocal plane. This gives $x/(a^2\ell)+y/(b^2m)+z/(c^2n)+1/p=0$.
5. Conclude. Verify that substituting the normal parametrisation for $P',Q',R'$ satisfies this plane equation.

Worked Example

2022 Paper 1, 2022-P1-Q2c (20 marks)

$P,Q,R;\,P',Q',R'$ are feet of the six normals from a point to the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$. The plane $PQR$ is $\ell x+my+nz=p$. Show the plane $P'Q'R'$ is $\dfrac{x}{a^2\ell}+\dfrac{y}{b^2m}+\dfrac{z}{c^2n}+\dfrac{1}{p}=0$.

Step 1 — Parametrise normal feet. From external point $(\alpha,\beta,\gamma)$, the foot of a normal at parameter $t$ is: $x_0=\frac{\alpha a^2}{a^2+t},\quad y_0=\frac{\beta b^2}{b^2+t},\quad z_0=\frac{\gamma c^2}{c^2+t}.$ Substituting into the ellipsoid gives a sextic in $t$ with six roots $t_1,\ldots,t_6$.

Step 2 — Plane $PQR$ gives a cubic. Three feet on $\ell x+my+nz=p$: $\frac{\ell\alpha}{a^2+t}+\frac{m\beta}{b^2+t}+\frac{n\gamma}{c^2+t}=p.\qquad(\ddagger)$ Clearing denominators gives a cubic in $t$ with roots $t_P,t_Q,t_R$.

Step 3 — The complementary plane. Consider the plane $\Pi'$: $\frac{x}{a^2\ell}+\frac{y}{b^2m}+\frac{z}{c^2n}+\frac{1}{p}=0.$ Substituting the normal parametrisation into $\Pi'$: $\frac{\alpha}{l(a^2+t)}+\frac{\beta}{m(b^2+t)}+\frac{\gamma}{n(c^2+t)}+\frac{1}{p}=0.\qquad(\dagger)$ This is a different cubic in $t$.

Step 4 — Factorisation argument. Equations $(\ddagger)$ and $(\dagger)$ are cubics whose product (after clearing denominators) equals the sextic from Step 1 (this is the classical Salmon factorisation). Hence the three roots of $(\dagger)$ are exactly $t_{P'},t_{Q'},t_{R'}$ — the complementary three feet.

Step 5 — Conclusion. Since $P',Q',R'$ correspond to values $t_{P'},t_{Q'},t_{R'}$ that satisfy $(\dagger)$, their coordinates satisfy $\Pi'$: $\boxed{\;\frac{x}{a^2\ell}+\frac{y}{b^2m}+\frac{z}{c^2n}+\frac{1}{p}=0.\;}\qquad\blacksquare$

Common Traps

• The reciprocal transformation is $\ell\to 1/(a^2\ell)$, $m\to 1/(b^2m)$, $n\to 1/(c^2n)$, $p\to -1/p$. The sign flip on $p$ is essential; it gives the $+1/p$ in the constant term of the reciprocal plane (not $-1/p$).
• This is a classical result. UPSC expects you to set up the parametrisation, write down equation $(\ddagger)$ for the first cubic, and explain (at a sketch-proof level) why the complementary cubic gives the reciprocal plane. Full algebraic verification of the Salmon factorisation is not expected in 20 marks.
• Do not confuse the six feet (which lie on the ellipsoid) with the six normals (which are lines from an external point). Each normal touches the ellipsoid at exactly one foot.
• The planes $PQR$ and $P'Q'R'$ are called conjugate planes of the foot-locus; they arise from a fundamental duality of the six-normals configuration.

Marks-Aware Writing

10-mark question (tangent planes through a line): Set up the pencil in two lines, write the tangency condition, solve for $\lambda$ (one quadratic), write the two planes. Verification is expected: show each plane contains the line and satisfies the tangency condition. Total: about 8 computation lines plus two verification checks.

15-mark question (normal chord + locus): Two parts — first derive the chord length $2S^{3/2}/S'$ (requires setting up the parametrisation, the quadratic in $t$, and reading off the second root), then impose $PP'=4PG_3$. Each part is worth roughly half the marks. Show the final locus is homogeneous of degree 2 (i.e., a cone) as a closing remark.

20-mark question (six normals, reciprocal plane): Structure: (1) parametrise the normal foot as $(x_0,y_0,z_0)$ in terms of $t$ — 4 marks; (2) derive the cubic $(\ddagger)$ for plane $PQR$ — 6 marks; (3) write the reciprocal plane and verify it corresponds to the complementary cubic $(\dagger)$ — 6 marks; (4) conclude — 4 marks. At this marks level, a clean statement of the Salmon factorisation principle with a coherent algebraic structure earns full marks even if the explicit sextic polynomial is not expanded.

Practice Set

• 2018-P2-Q5a (10 m) — — Hint: tangent plane or normal to an ellipsoid; read off $a^2,b^2,c^2$ first.
• 2017-P1-Q3d (10 m) — — Hint: director sphere — sum orthonormal tangency conditions.
• 2016-P1-Q4d (15 m) — — Hint: director sphere at 15 marks — state orthonormality identities explicitly.