Hyperboloid of one sheet

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2013, 2014, 2026)
• Priority tier: T3
• Marks (count): 15 (1), 20 (1)
• Average solve time: ~22 min
• Difficulty mix: hard 1, medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

The hyperboloid of one sheet is the canonical example of a doubly-ruled surface: every point has two lines of each ruling family passing through it. UPSC tests exactly two things about it: (1) finding the equations of the two generators through a specific point, and (2) proving an invariant product relation involving a variable generator and fixed generators through the minor-axis endpoints. Both questions reduce to the same fundamental technique — the two-family factorisation. Master the factorisation and the parameter-substitution method and you can handle both archetypes at any marks level.

Minimum Theory

The hyperboloid of one sheet. The standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1.$ The principal elliptic section is the curve $z=0$: $x^2/a^2+y^2/b^2=1$, an ellipse with semi-axes $a$ (along $x$) and $b$ (along $y$). The minor axis of this ellipse has endpoints $B=(0,b,0)$ and $B'=(0,-b,0)$.

Two-family factorisation. Rewrite the hyperboloid as $\left(\frac{x}{a}-\frac{z}{c}\right)\!\!\left(\frac{x}{a}+\frac{z}{c}\right)=\left(1-\frac{y}{b}\right)\!\!\left(1+\frac{y}{b}\right).$ Assign one factor on each side to define two families of lines:

$\textbf{Family }\Lambda\;(\text{parameter }\lambda):\quad \frac{x}{a}-\frac{z}{c}=\lambda\!\left(1-\frac{y}{b}\right),\quad\frac{x}{a}+\frac{z}{c}=\frac{1}{\lambda}\!\left(1+\frac{y}{b}\right).$

$\textbf{Family }M\;(\text{parameter }\mu):\quad \frac{x}{a}-\frac{z}{c}=\mu\!\left(1+\frac{y}{b}\right),\quad\frac{x}{a}+\frac{z}{c}=\frac{1}{\mu}\!\left(1-\frac{y}{b}\right).$

Every point on the hyperboloid lies on exactly one generator of each family. Two generators of the same family never intersect (they are skew); a generator of family $\Lambda$ always meets every generator of family $M$.

Direction of a generator. For a generator of family $\Lambda$ at parameter $\lambda$, the direction vector is $(-a\sin\theta,\;b\cos\theta,\;c)$ at the point $(a\cos\theta,b\sin\theta,0)$ on the principal section. Family $M$ gives direction $(-a\sin\theta,\;b\cos\theta,\;-c)$ at the same point — the $z$-component flips.

Question Archetypes

generating-lines-through-point (1 question(s); 2014)

Recognition Cues

• “Find the equations of the two generating lines through the point $(a\cos\theta,b\sin\theta,0)$ of the principal elliptic section.”
• A specific point on the hyperboloid is given; you need the line equations, not just the direction.
• The two generators share the same $(x,y)$-direction (tangent to the ellipse at the point) but differ in the sign of the $z$-component.

Solution Template

1. Write the factorisation. State the hyperboloid in factored form $(x/a-z/c)(x/a+z/c)=(1-y/b)(1+y/b)$.
2. Define the two families $\Lambda$ and $M$ with parameters $\lambda$ and $\mu$.
3. Substitute the given point $(x,y,z)=(a\cos\theta,b\sin\theta,0)$ into the first equation of each family to find $\lambda$ and $\mu$.
4. Find the direction vector. For family $\Lambda$: differentiate/parametrise along the generator or use the two-family equations at the given point to read off the direction. The direction at the principal section point simplifies to $(-a\sin\theta,\;b\cos\theta,\;\pm c)$.
5. Write the line equations through $(a\cos\theta,b\sin\theta,0)$ with the computed directions.

Worked Example

2014 Paper 1, 2014-P1-Q4c (15 marks)

Find the equations of the two generating lines through any point $(a\cos\theta,\,b\sin\theta,\,0)$ of the principal elliptic section $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1,\;z=0$ of the hyperboloid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1$.

Step 1 — Factorisation. $\left(\frac{x}{a}-\frac{z}{c}\right)\!\!\left(\frac{x}{a}+\frac{z}{c}\right)=\left(1-\frac{y}{b}\right)\!\!\left(1+\frac{y}{b}\right).$

Step 2 — Two families.

Family $\Lambda$: $\quad\dfrac{x}{a}-\dfrac{z}{c}=\lambda\!\left(1-\dfrac{y}{b}\right)$, $\quad\dfrac{x}{a}+\dfrac{z}{c}=\dfrac{1}{\lambda}\!\left(1+\dfrac{y}{b}\right)$.

Family $M$: $\quad\dfrac{x}{a}-\dfrac{z}{c}=\mu\!\left(1+\dfrac{y}{b}\right)$, $\quad\dfrac{x}{a}+\dfrac{z}{c}=\dfrac{1}{\mu}\!\left(1-\dfrac{y}{b}\right)$.

Step 3 — Find $\lambda,\mu$. At $P=(a\cos\theta,b\sin\theta,0)$: $x/a=\cos\theta$, $y/b=\sin\theta$, $z/c=0$.

From family $\Lambda$: $\cos\theta=\lambda(1-\sin\theta)\;\Rightarrow\;\lambda=\dfrac{\cos\theta}{1-\sin\theta}$.

From family $M$: $\cos\theta=\mu(1+\sin\theta)\;\Rightarrow\;\mu=\dfrac{\cos\theta}{1+\sin\theta}$.

(Both second equations are satisfied: $\cos\theta/\lambda=(1+\sin\theta)$ follows from $\cos^2\theta=(1-\sin\theta)(1+\sin\theta)$.)

Step 4 — Direction vectors. Using $1-\lambda^2=-2\sin\theta/(1-\sin\theta)$ and $1+\lambda^2=2/(1-\sin\theta)$:

Family $\Lambda$: $\vec d_\Lambda\propto(-a\sin\theta,\;b\cos\theta,\;c)$.

Family $M$: $\vec d_M\propto(-a\sin\theta,\;b\cos\theta,\;-c)$.

Step 5 — Line equations.

$\boxed{\;\text{Generator }\Lambda:\quad\frac{x-a\cos\theta}{-a\sin\theta}=\frac{y-b\sin\theta}{b\cos\theta}=\frac{z}{c}.\;}$

$\boxed{\;\text{Generator }M:\quad\frac{x-a\cos\theta}{-a\sin\theta}=\frac{y-b\sin\theta}{b\cos\theta}=\frac{z}{-c}.\;}$

The two generators share the same $(x,y)$-direction (tangent to the principal ellipse at $P$) and are reflections of each other across the $z=0$ plane.

Common Traps

• Both generators share the $(x,y)$-direction $(-a\sin\theta,b\cos\theta)$, which is the tangent to the ellipse $x^2/a^2+y^2/b^2=1$ at $(\cos\theta,\sin\theta)$. Only the $z$-component changes sign.
• The second equation of each family is automatic. If the first family $\Lambda$ equation gives $\lambda=\cos\theta/(1-\sin\theta)$, verify the second using $\cos^2\theta=(1-\sin\theta)(1+\sin\theta)$ — no separate calculation is needed.
• The two families are not symmetric in $y$: family $\Lambda$ has $(1-y/b)$ on the right of the first equation and family $M$ has $(1+y/b)$. Swapping them exchanges the families.
• At $\theta=\pi/2$ (i.e., $P=B=(0,b,0)$): the formula gives $\sin\theta=1$, $1-\sin\theta=0$, so $\lambda\to\infty$. At this special point, the family $\Lambda$ generator reduces to $y=b,\;x/a=z/c$, and the family $M$ generator to $y=b,\;x/a=-z/c$ — read them directly from the factorised form.

generator-product-property (1 question(s); 2013)

Recognition Cues

• “A variable generator meets the fixed generators through the endpoints $B$ and $B'$ of the minor axis in $P$ and $P'$; prove $BP\cdot B'P'=a^2+c^2$.”
• You need to compute distances from $B$ to $P$ and from $B'$ to $P'$, then show their product is a constant.
• The product is independent of the parameter $\lambda$ of the variable generator.

Solution Template

1. Identify $B,B'$ and their generators. $B=(0,b,0)$ lies on family $M$ generator $L_B$ with $y=b$, $x/a=z/c$. $B'=(0,-b,0)$ lies on family $M$ generator $L_{B'}$ with $y=-b$, $x/a=-z/c$.
2. Take a variable generator from family $\Lambda$ at parameter $\lambda$. Find its intersection $P$ with $L_B$ by setting $y=b$ and solving the family $\Lambda$ equations.
3. Find intersection $P'$ with $L_{B'}$ by setting $y=-b$.
4. Compute $BP$ and $B'P'$. Find $\overrightarrow{BP}=P-B$ and $\overrightarrow{B'P'}=P'-B'$; compute their magnitudes.
5. Show $BP\cdot B'P'=a^2+c^2$ by noting that $|\lambda|$ and $1/|\lambda|$ cancel in the product.

Worked Example

2013 Paper 1, 2013-P1-Q4c (20 marks)

A variable generator meets two generators of the system through the extremities $B$ and $B'$ of the minor axis of the principal elliptic section of $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-\dfrac{z^2}{c^2}=1$ in $P$ and $P'$. Prove that $BP\cdot B'P'=a^2+c^2$.

Step 1 — Generators through $B$ and $B'$.

$B=(0,b,0)$: substitute $y=b$ ($\Rightarrow 1+y/b=2$, $1-y/b=0$) into the family $M$ factorisation. The generator through $B$ is $L_B: \{y=b,\;x/a=z/c\}$, with direction $(a,0,c)$.

$B'=(0,-b,0)$: substitute $y=-b$ ($\Rightarrow 1+y/b=0$, $1-y/b=2$) into family $M$. The generator through $B'$ is $L_{B'}: \{y=-b,\;x/a=-z/c\}$, with direction $(a,0,-c)$.

Both $L_B$ and $L_{B'}$ are in family $M$. They are skew (different $y$-values, non-parallel directions).

Step 2 — Variable generator $\ell_\lambda$ from family $\Lambda$. At parameter $\lambda$: $\frac{x}{a}-\frac{z}{c}=\lambda\!\left(1-\frac{y}{b}\right),\quad\frac{x}{a}+\frac{z}{c}=\frac{1}{\lambda}\!\left(1+\frac{y}{b}\right).$

Step 3 — Intersection $P=\ell_\lambda\cap L_B$. Set $y=b$ (so $1-y/b=0$, $1+y/b=2$): $x/a-z/c=0\quad\text{and}\quad x/a+z/c=2/\lambda.$ Adding: $2x/a=2/\lambda\Rightarrow x=a/\lambda$. Then $z=c/\lambda$. $P=\bigl(a/\lambda,\;b,\;c/\lambda\bigr).$

Step 4 — Intersection $P'=\ell_\lambda\cap L_{B'}$. Set $y=-b$ (so $1-y/b=2$, $1+y/b=0$): $x/a-z/c=2\lambda\quad\text{and}\quad x/a+z/c=0.$ Adding: $2x/a=2\lambda\Rightarrow x=a\lambda$. Then $z=-c\lambda$. $P'=\bigl(a\lambda,\;-b,\;-c\lambda\bigr).$

Step 5 — Distances.

$\overrightarrow{BP}=P-B=\bigl(a/\lambda,\;0,\;c/\lambda\bigr)\;\Longrightarrow\;BP=\sqrt{\frac{a^2}{\lambda^2}+\frac{c^2}{\lambda^2}}=\frac{\sqrt{a^2+c^2}}{|\lambda|}.$

$\overrightarrow{B'P'}=P'-B'=\bigl(a\lambda,\;0,\;-c\lambda\bigr)\;\Longrightarrow\;B'P'=\sqrt{a^2\lambda^2+c^2\lambda^2}=|\lambda|\sqrt{a^2+c^2}.$

Step 6 — Product. $BP\cdot B'P'=\frac{\sqrt{a^2+c^2}}{|\lambda|}\cdot|\lambda|\sqrt{a^2+c^2}=a^2+c^2.$

$\boxed{\;BP\cdot B'P'=a^2+c^2.\;}\qquad\blacksquare$

The $|\lambda|$ and $1/|\lambda|$ factors cancel exactly — this is the geometric invariant. $\checkmark$

Verification. Take $a=b=c=1$, $\lambda=1$: $P=(1,1,1)$, $B=(0,1,0)$, $BP=\sqrt{2}$. $P'=(1,-1,-1)$, $B'=(0,-1,0)$, $B'P'=\sqrt{2}$. Product $=2=a^2+c^2$ ✓.

Common Traps

• Both $L_B$ and $L_{B'}$ must come from the same ruling family. They are both in family $M$; the variable generator comes from family $\Lambda$. A generator of $\Lambda$ meets every generator of $M$, so both intersections exist. Two generators of the same family are skew and do not intersect.
• The $|\lambda|$ cancellation is the entire point. Before computing the product, observe that $BP\propto 1/|\lambda|$ and $B'P'\propto|\lambda|$; the product must be $|\lambda|$-free. If your intermediate distances don’t show this structure, recheck the intersection calculation.
• PDF transcription note. The 2013 paper as printed shows "$z^2c^2$" in the hyperboloid equation; the standard correct form is $z^2/c^2$. The answer $a^2+c^2$ (not $a^2+1/c^2$) is consistent with the correct form.
• $B=(0,b,0)$ is an endpoint of the minor axis of the principal ellipse only when $a>b$. If $a<b$, the endpoints $(0,\pm b,0)$ are the major-axis endpoints, but the calculation is identical — the problem label “minor” refers to the $b$-axis regardless.

Marks-Aware Writing

15-mark question (generating lines through a point): Show the factorisation in full (2 marks), define both families (2 marks), find $\lambda$ and $\mu$ by substitution (4 marks), derive the direction vectors (4 marks), write the two symmetric line equations (3 marks). A verification that each line equation is satisfied at $P$ and that the direction lies on the tangent plane is a cheap closing check.

20-mark question (generator product property): Allocate roughly: factorisation + families (4 marks), identify $L_B$ and $L_{B'}$ with justification that they are same-family and skew (4 marks), find $P$ and $P'$ by explicit substitution (6 marks), compute $BP$ and $B'P'$ (4 marks), compute and interpret the product (2 marks). The key sentence for full marks: “The factor $|\lambda|$ in $B'P'$ exactly cancels the $1/|\lambda|$ in $BP$, making the product independent of $\lambda$.”

Practice Set

• 2020-P1-Q4b (15 m) — — Hint: ruled surface or hyperboloid — use the two-family factorisation.