Paraboloid (elliptic and hyperbolic)

At a Glance

• Frequency: 6 sub-parts across 6 of 13 years (2015, 2017, 2018, 2019, 2020, 2023)
• Priority tier: T2
• Marks (count): 10 (2), 13 (2), 15 (2)
• Average solve time: ~11 min
• Difficulty mix: easy 2, hard 4
• Section: A | Dominant type: derivation

Why This Chapter Matters

Paraboloid questions appear in 6 of the last 13 years and are the most technically demanding atom in Paper 1 Analytic Geometry. Four of the six questions are rated hard. The atom covers a wide range: a simple tangent-plane gradient calculation (2017), completing the square to identify canonical form (2023), finding generating lines of a ruled quadric (2018), the locus of perpendicular generators (2020), a cubic normal-count condition (2019), and an envelope-of-tangent-planes problem (2015). The first two are 10-mark easy questions handled by standard gradient/completion-of-squares; the remaining four require deeper technique and care.

Minimum Theory

Canonical forms.

Tangent plane. At a point $(x_0,y_0,z_0)$ on the surface $F(x,y,z)=0$, the tangent plane is $\nabla F\cdot(\mathbf r-\mathbf r_0)=0$.

For the elliptic paraboloid $x^2/a^2+y^2/b^2-z=0$: tangent at $(x_0,y_0,z_0)$ is $\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=\frac{z+z_0}{2}\qquad\text{(split-equation rule)}.$ For the general paraboloid $\alpha x^2+\beta y^2=2z$: tangent at $(\xi,\eta,\zeta)$ is $\alpha\xi x+\beta\eta y=z+\zeta$.

Reduction to canonical form. Complete the square in $x$ and $y$ to eliminate cross-terms. Translate coordinates to move the vertex to the origin.

Generating lines (rulings) of the hyperbolic paraboloid. Factor $x^2/a^2-y^2/b^2=2z$ as $\left(\frac{x}{a}-\frac{y}{b}\right)\!\left(\frac{x}{a}+\frac{y}{b}\right)=2z.$ Two families of rulings (each pair of matching $\lambda/\mu$ gives a line lying entirely on the surface):

• $\lambda$-family: $\dfrac{x}{a}-\dfrac{y}{b}=2\lambda$ and $\lambda\!\left(\dfrac{x}{a}+\dfrac{y}{b}\right)=z$.
• $\mu$-family: $\dfrac{x}{a}+\dfrac{y}{b}=2\mu$ and $\mu\!\left(\dfrac{x}{a}-\dfrac{y}{b}\right)=z$.

Finding generators through a point. For a ruled quadric at a surface point $(x_0,y_0,z_0)$, parametrise the line as $(x_0+at,y_0+bt,z_0+ct)$; substitute; the $t^0$ term vanishes (point is on surface); impose $t^1=0$ and $t^2=0$ to get two equations in $(a:b:c)$ — these factor into two directions.

Normals to a paraboloid. The normal at $(\xi,\eta,\zeta)$ on $x^2+y^2=2az$ passes through a given point $(\alpha,\beta,\gamma)$ iff a cubic equation in $\lambda$ (the parameter along the normal) is satisfied — generically three normals.

Envelope of a family of planes. For a 1-parameter family of planes $F(x,y,z,t)=0$: the envelope is the curve/surface obtained by eliminating $t$ from $F=0$ and $\partial F/\partial t=0$.

Question Archetypes

Six patterns cover all six questions in the corpus.

tangent-plane-at-point (1 question(s); 2017)

Recognition Cues

• “Find the equation of the tangent plane at point $(x_0,y_0,z_0)$ to the conicoid $F(x,y,z)=0$.”

Solution Template

1. Verify the point lies on the surface: $F(x_0,y_0,z_0)=0$.
2. Compute $\nabla F$ and evaluate at $(x_0,y_0,z_0)$.
3. Write the tangent plane: $\nabla F\cdot(\mathbf r-\mathbf r_0)=0$.
4. Simplify and double-check using the split-equation shortcut.

Worked Example(s)

2017 Paper 1, 2017-P1-Q1d (10 marks, compulsory)

Find the tangent plane at $(1,1,1)$ to $3x^2-y^2=2z$.

$F=3x^2-y^2-2z$. Check: $3-1-2=0$ ✓. $\nabla F=(6x,-2y,-2)$; at $(1,1,1)$: $(6,-2,-2)$.

Tangent plane: $6(x-1)-2(y-1)-2(z-1)=0\Rightarrow 6x-2y-2z=2$: $\boxed{\;3x-y-z=1.\;}$

Common Traps

• Keep track of the sign on the $z$-term: $F=3x^2-y^2-2z$ gives $F_z=-2$, not $+2$.
• The split-equation shortcut for $\alpha x^2+\beta y^2=2z$: replace each $x^2\to xx_0$, $y^2\to yy_0$, $z\to(z+z_0)/2$. Here: $3(x)(1)-(y)(1)=(z+1)$, i.e., $3x-y-z=1$ ✓.

paraboloid-reduction (1 question(s); 2023)

Recognition Cues

• “Show that $\alpha x^2+\beta y^2+\ldots=0$ represents an elliptic/hyperbolic paraboloid.”
• “Find its vertex, principal axis and principal planes.”

Solution Template

1. Complete the square in $x$ and in $y$.
2. Move all constant terms to the right side; the equation should take the form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=z-z_0$ (elliptic) or with a minus sign (hyperbolic).
3. Read off vertex $(h,k,z_0)$, axis (line through vertex parallel to $z$-axis), principal planes (the two planes of symmetry through the axis).

Worked Example(s)

2023 Paper 1, 2023-P1-Q2c-i (10 marks)

Show $2x^2+3y^2-8x+6y-12z+11=0$ is an elliptic paraboloid. Find its principal axis and planes.

Complete the square: $2(x-2)^2-8+3(y+1)^2-3-12z+11=0\;\Rightarrow\;2(x-2)^2+3(y+1)^2=12z.$ Divide by 12: $\dfrac{(x-2)^2}{6}+\dfrac{(y+1)^2}{4}=z$.

This is canonical form — elliptic paraboloid with vertex $(2,-1,0)$. $\boxed{\;\text{Principal axis: }x=2,\;y=-1.\quad\text{Principal planes: }x=2\text{ and }y=-1.\;}$

Common Traps

• Track the signs on the completing-the-square constants carefully: $2(x-2)^2=2x^2-8x+8$, so the residual is $-8+3=$ etc.
• The “principal planes” for a paraboloid are the two planes of symmetry through the axis — each is parallel to one coordinate plane and contains the axis.

generating-lines (1 question(s); 2018)

Recognition Cues

• “Find the equations of the generating lines of [ruled quadric] through point $P$.”
• The surface equation is often presented as a product of two linear forms.

Solution Template

1. Verify $P$ is on the surface.
2. Parametrise: $(x,y,z)=P+(a,b,c)t$.
3. Substitute into $F(x,y,z)=0$; expand as a polynomial in $t$.
4. The $t^0$ coefficient vanishes (P on surface). Set $t^1=0$ (linear condition on $(a,b,c)$) and $t^2=0$ (quadratic condition).
5. From the linear condition express $c$ in terms of $a,b$; substitute into the quadratic to get a factored equation in $(a:b)$; two factors give two direction ratios.

Worked Example(s)

2018 Paper 1, 2018-P1-Q3c (13 marks)

Find the generating lines of $(x+y+z)(2x+y-z)=6z$ through $(1,1,1)$.

Verify: $(1+1+1)(2+1-1)=3\cdot2=6=6\cdot1$ ✓.

Parametrise $(1+at,1+bt,1+ct)$. Substituting, the $t^1$ coefficient gives: $(\text{L})\quad 8a+5b-7c=0\;\Rightarrow\;c=\frac{8a+5b}{7}.$ The $t^2$ coefficient gives $2a^2+3ab+ac+b^2-c^2=0$. Substituting $c$ and multiplying by $49$: $6(3a+b)(5a+4b)=0.$ Two cases:

• $b=-3a$: take $a=1$, $c=-1$. Line: $\dfrac{x-1}{1}=\dfrac{y-1}{-3}=\dfrac{z-1}{-1}$.
• $b=-\tfrac{5}{4}a$: take $a=4$, $c=1$. Line: $\dfrac{x-1}{4}=\dfrac{y-1}{-5}=\dfrac{z-1}{1}$.

$\boxed{\;\frac{x-1}{1}=\frac{y-1}{-3}=\frac{z-1}{-1}\quad\text{and}\quad\frac{x-1}{4}=\frac{y-1}{-5}=\frac{z-1}{1}.\;}$

Verification: substitute $(1+t,1-3t,1-t)$ into $(x+y+z)(2x+y-z)-6z$: $(3-3t)(2)-6(1-t)=6-6t-6+6t=0$ ✓.

Common Traps

• The surface must be a ruled quadric (product of two linear forms or similar structure) for generating lines to exist.
• The $t^0$ term vanishes automatically; missing the $t^1=0$ condition means you only impose $t^2=0$ and find only one equation — not two directions.
• The quadratic in $(a,b)$ always factors for a ruled quadric; if you can’t factor it, recheck arithmetic.

normals-from-point (1 question(s); 2019)

Recognition Cues

• “Prove that three normals can be drawn from a given point to the paraboloid $x^2+y^2=2az$.”
• “If the point lies on the surface $27a(x^2+y^2)+8(a-z)^3=0$, two normals coincide.”

Solution Template

1. Write the normal direction at a surface point $(\xi,\eta,\zeta)$: for $x^2+y^2-2az=0$ it is $(\xi,\eta,-a)$.
2. Parametrise the normal through the external point $(\alpha,\beta,\gamma)$: write $(\xi,\eta,\zeta)$ in terms of a parameter $\lambda$.
3. Impose that the foot lies on the surface: this gives a cubic in $\lambda$ (three roots = three normals).
4. The cubic has a repeated root (two normals coincide) iff its discriminant $\Delta=0$; compute and identify the algebraic surface $\Delta=0$.

Worked Example(s)

2019 Paper 1, 2019-P1-Q3b (15 marks)

Prove three normals can be drawn from $(\alpha,\beta,\gamma)$ to $x^2+y^2=2az$; find the coincidence-condition surface.

Normal at $(\xi,\eta,\zeta)$ has direction $(\xi,\eta,-a)$. The foot satisfies $\xi=\alpha/(1-\lambda)$, $\eta=\beta/(1-\lambda)$, $\zeta=\gamma-a\lambda$. Substituting into $\xi^2+\eta^2=2a\zeta$: $\frac{\alpha^2+\beta^2}{(1-\lambda)^2}=2a(\gamma-a\lambda).$ Expanding, this is a cubic in $\lambda$ with leading coefficient $2a^2\ne0$ — three roots in general.

A repeated root requires $\Delta=0$. Computing and factoring the discriminant: $\Delta\;\propto\;8(a-\gamma)^3+27a(\alpha^2+\beta^2)=0,$ which (replacing $(\alpha,\beta,\gamma)\to(x,y,z)$) is: $\boxed{\;27a(x^2+y^2)+8(a-z)^3=0.\;}$

Common Traps

• The normal direction at $(\xi,\eta,\zeta)$ on $x^2+y^2-2az=0$ is $(\xi,\eta,-a)$ (not $(\xi,\eta,-2a)$ — the gradient gives $(2\xi,2\eta,-2a)$, which is proportional to $(\xi,\eta,-a)$).
• The derivation of the cubic requires careful expansion; confirm the leading coefficient is $2a^2$ (not zero) to assert it genuinely is a cubic.
• The discriminant carries spurious factors; discard $a^3$ and $(\alpha^2+\beta^2)$ (which zero out only in degenerate cases) to isolate the geometric condition.

perpendicular-generators-locus (1 question(s); 2020)

Recognition Cues

• “Find the locus of the point of intersection of perpendicular generators of the hyperbolic paraboloid $x^2/a^2-y^2/b^2=2z$.”

Solution Template

1. Write down the two families of rulings using the factored form.
2. Find the direction vectors of a $\lambda$-ruling and a $\mu$-ruling (cross-product of the two plane normals for each family).
3. Impose perpendicularity: $\mathbf d_1\cdot\mathbf d_2=0$; express as a condition on $\lambda\mu$.
4. At the intersection point, express $z=2\lambda\mu$ using the ruling equations; substitute the perpendicularity condition.

Worked Example(s)

2020 Paper 1, 2020-P1-Q4b (15 marks)

Locus of intersection of perpendicular generators of $x^2/a^2-y^2/b^2=2z$.

$\lambda$-generator direction (after cross-product of its plane normals, scaled by $ab$): $\mathbf d_1=(a,b,2\lambda)$.

$\mu$-generator direction: $\mathbf d_2=(-a,b,-2\mu)$.

Perpendicularity: $\mathbf d_1\cdot\mathbf d_2=-a^2+b^2-4\lambda\mu=0\;\Rightarrow\;4\lambda\mu=b^2-a^2$.

$z$-coordinate of intersection: From the ruling equations, $z=\lambda\cdot(x/a+y/b)=\lambda\cdot2\mu=2\lambda\mu=(b^2-a^2)/2$.

The locus is the plane: $\boxed{\;2z=b^2-a^2.\;}$

Common Traps

• The two perpendicular generators must come from different families ($\lambda$- and $\mu$-families), not from the same family.
• The direction vectors scale by $ab$; include this factor consistently in both $\mathbf d_1$ and $\mathbf d_2$ before taking the dot product.
• The locus is a constant-$z$ plane, not a curve; $x$ and $y$ remain free.

tangent-plane-envelope (1 question(s); 2015)

Recognition Cues

• “Two perpendicular tangent planes to [paraboloid] intersect in a straight line in the plane $x=0$. Obtain the curve this line touches.”
• The answer is the envelope of a 1-parameter family of lines.

Solution Template

1. Write the tangent-plane equation at a general point $(\alpha,\beta)$: $\alpha x+\beta y-z=(\alpha^2+\beta^2)/2$ for $x^2+y^2=2z$.
2. Perpendicularity of two such planes: $\alpha_1\alpha_2+\beta_1\beta_2+1=0$.
3. Line in $x=0$: both planes restricted to $x=0$ must coincide, forcing $\beta_1=\beta_2=\beta$, $\alpha_2=-\alpha_1$. Perpendicularity then gives $\beta^2-\alpha^2=-1$.
4. Write the 1-parameter family of lines $F(y,z,\beta)=0$; find the envelope by eliminating $\beta$ from $F=0$ and $\partial F/\partial\beta=0$.

Worked Example(s)

2015 Paper 1, 2015-P1-Q4c (13 marks)

Two perpendicular tangent planes to $x^2+y^2=2z$ intersect in a line in $x=0$. Find the curve this line touches.

Tangent plane at $(\alpha,\beta,(\alpha^2+\beta^2)/2)$: $\alpha x+\beta y-z=(\alpha^2+\beta^2)/2$.

For two planes to intersect along a line in $x=0$: $\beta_1=\beta_2=\beta$, $\alpha_2=-\alpha$. Perpendicularity: $-\alpha^2+\beta^2=-1$, so $\alpha^2-\beta^2=1$.

On $x=0$ both planes give $\beta y-z=(\alpha^2+\beta^2)/2=(1+2\beta^2)/2$, i.e.: $F(y,z,\beta)=z-\beta y+\beta^2+\tfrac{1}{2}=0.$

Envelope: $\partial F/\partial\beta=-y+2\beta=0\Rightarrow\beta=y/2$. Substitute: $z=\tfrac{y^2}{2}-\tfrac{y^2}{4}-\tfrac{1}{2}=\tfrac{y^2}{4}-\tfrac{1}{2}.$ $\boxed{\;x=0,\quad y^2=4z+2.\;}$

Common Traps

• “Intersect in a line in $x=0$” forces the traces of both planes on $x=0$ to coincide. This gives $\beta_1=\beta_2$ and $\alpha_2=-\alpha_1$ — both conditions must be derived, not guessed.
• In the perpendicularity condition, the normals of the tangent planes are $(\alpha,-(-\beta),-1)=(\alpha,\beta,-1)$; the dot product includes the $(-1)(-1)=+1$ from the $z$-components.
• The envelope step is $F=0$ and $F_\beta=0$ simultaneously — don’t confuse $F_\beta$ with $\partial F/\partial y$.

Marks-Aware Writing

10-mark questions (2017, 2023): Tangent plane: three lines (verify point, gradient, equation). Paraboloid reduction: display the completing-the-square step clearly, state the canonical form, identify vertex/axis/planes.

13-mark questions (2015, 2018): Show all intermediate steps — the factoring of the quadratic condition (2018) or the envelope derivation (2015) is worth 5–6 marks on its own.

15-mark questions (2019, 2020): These need systematic setup. For normals: derive the cubic, state its degree and leading coefficient, then address the discriminant. For perpendicular generators: write the ruling families, direction vectors, dot product — each step is a mark.

Practice Set