Quadric surfaces ellipsoid hyperboloid paraboloid: 28 Practice MCQs with Answers and Explanation

Q1. A student derives the equation $z = x^2 - y^2$ and claims the surface is an elliptic paraboloid because it contains squared terms. Which analysis correctly identifies the error in this classification?

A.The student failed to complete the square; completing the square reveals a hyperbolic structure.

B.The presence of opposite signs on squared variables defines a hyperbolic paraboloid, not an elliptic one, regardless of the linear z term. ✅

C.The student confused the axis of symmetry; rotating coordinates would yield positive coefficients for both squares.

D.An elliptic paraboloid requires a constant term on the right side, which is missing here.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This question targets error analysis by addressing the common misconception that any quadratic form with a linear variable is elliptic. Students must recognize that the sign difference between $x^2$ and $y^2$ fundamentally dictates the saddle geometry of a hyperbolic paraboloid, distinguishing it from the bowl shape of elliptic surfaces.

Q2. When modeling a cooling tower for a power plant, engineers require a doubly ruled surface that can be constructed using straight steel beams while maintaining structural integrity under vertical loads. Which quadric surface best satisfies these geometric and physical constraints?

A.Ellipsoid

B.Hyperboloid of One Sheet ✅

C.Elliptic Paraboloid

D.Cone

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This application-based scenario requires connecting abstract geometric properties to real-world engineering needs. The hyperboloid of one sheet is uniquely suited because it is a doubly ruled surface, allowing straight beams to form a curved structure. Other options either lack the ruling property or are singular at the origin, making them structurally inferior for towers.

Q3. Consider the family of surfaces defined by $x^2 + y^2 - z^2 = k$ . As the parameter $k$ transitions from positive to negative values through zero, how does the topology of the level sets change?

A.The surface remains connected but changes from opening along the z-axis to opening radially.

B.The surface splits into two disconnected sheets when

k > 0

and merges into one sheet when

k < 0

.

C.The surface is a single connected component for

k > 0

, becomes a cone at

k=0

, and splits into two disconnected sheets for

k < 0

. ✅

D.The topology remains invariant; only the scaling factor changes proportionally to

|k|

.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This mixed-concept question demands understanding parametric families and topological connectivity. Students must visualize the transition: $k>0$ yields a hyperboloid of one sheet (connected), $k=0$ yields a cone (singular connected), and $k<0$ yields a hyperboloid of two sheets (disconnected). This tests dynamic spatial reasoning beyond static identification.

Q4. A contour map shows level curves $z = c$ that are hyperbolas for all $c \neq 0$ and a pair of intersecting lines at $c = 0$ . If the surface passes through the origin and is smooth everywhere except possibly at the origin, which equation could represent this surface?

A.

z = x^2 + y^2

B.

z^2 = x^2 + y^2

C.

z = x^2 - y^2

✅

D.

x^2 + y^2 + z^2 = 1

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Students must reverse-engineer the surface from cross-sectional data. Hyperbolic level curves indicate opposite signs in the quadratic form, pointing to $z = x^2 - y^2$ . The intersecting lines at $c=0$ confirm the saddle point at the origin. This graph-interpretation skill is essential for analyzing surfaces without explicit 3D visualization.

Q5. Which of the following transformations converts the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ into a unit sphere, and what is the Jacobian determinant of this transformation used in volume integration?

A.

u=x/a, v=y/b, w=z/c

; Jacobian is

abc

✅

B.

u=ax, v=by, w=cz

; Jacobian is

1/(abc)

C.

u=x/a, v=y/b, w=z/c

; Jacobian is

1/(abc)

D.

u=x^2/a^2, v=y^2/b^2, w=z^2/c^2

; Jacobian is

8abc\sqrt{uvw}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This multi-step problem links coordinate transformation to geometric scaling. Substituting $u=x/a$ etc. normalizes the ellipsoid to a unit sphere. The Jacobian accounts for volume distortion: since $x=au$ , $dx=a\,du$ , the differential volume scales by $abc$ . Confusing forward/inverse mappings leads to reciprocal errors, a common pitfall.

Q6. A student attempts to find the intersection of the paraboloid $z = x^2 + y^2$ and the plane $z = 2x + 2y$ . They conclude the intersection is empty because substituting gives $x^2 - 2x + y^2 - 2y = 0$ , which they claim has no real solutions. What is the flaw in this reasoning?

A.The student forgot that the equation represents a circle after completing the square, not an impossible condition. ✅

B.The student should have used cylindrical coordinates instead of Cartesian substitution.

C.The plane is tangent to the paraboloid, so the intersection is a single point, not empty.

D.The equation actually describes a hyperbola, which always has real solutions.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This error-analysis question exposes incomplete algebraic manipulation. Completing the square yields $(x-1)^2 + (y-1)^2 = 2$ , a valid circle. Students often misjudge quadratic forms without normalization. Recognizing that sum-of-squares equaling a positive constant implies real solutions is crucial for correctly analyzing intersections in 3D geometry.

Q7. In satellite dish design, signals reflect off a parabolic surface to a focal point. If the dish is modeled by $z = \frac{x^2 + y^2}{4p}$ , and manufacturing tolerances introduce a small cubic perturbation $\epsilon(x^3 + y^3)$ , how does this affect signal focusing compared to the ideal quadric?

A.The focus shifts along the z-axis but remains a single point due to rotational symmetry preservation.

B.Astigmatism occurs because the perturbation breaks rotational symmetry, causing focal points to spread into a caustic curve. ✅

C.The focal length increases uniformly, requiring only a receiver repositioning without quality loss.

D.Higher-order terms vanish at the vertex, so focusing remains perfect near the center regardless of

\epsilon

.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This Olympiad-style question blends perturbation theory with quadric optics. Cubic terms break the perfect rotational symmetry of the paraboloid, introducing aberrations. Unlike quadratic deformations that merely shift focus, odd-powered perturbations create asymmetric wavefront errors, leading to astigmatic foci. This tests deep understanding of why quadrics are uniquely optimal for focusing.

Q8. Given the general second-degree equation $Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fzx + Gx + Hy + Iz + J = 0$ , which invariant quantity determines whether the surface is central or non-central without diagonalizing the matrix?

A.The determinant of the full 4x4 augmented matrix

B.The trace of the 3x3 quadratic form matrix

C.The rank of the 3x3 quadratic submatrix relative to the augmented system ✅

D.The sign of the product

ABC

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This conceptual question addresses classification invariants. Central quadrics (ellipsoids, hyperboloids) have a unique center, occurring when the gradient system has a solution. This depends on consistency between the quadratic and linear parts, captured by comparing ranks. Determinant alone doesn't distinguish paraboloids (non-central) from degenerate cases, making rank analysis essential for proper categorization.

Q9. Two students analyze the surface $x^2 + 4y^2 - z^2 + 4x - 8y + 2z = 5$ . Student A completes the square and identifies it as a hyperboloid of one sheet centered at $(-2,1,1)$ . Student B claims it's an elliptic paraboloid because of mixed signs. Who is correct and why?

A.Student B is correct; mixed signs always indicate a paraboloid.

B.Student A is correct; completing the square yields

(x+2)^2 + 4(y-1)^2 - (z-1)^2 = 6

, confirming a hyperboloid of one sheet. ✅

C.Both are wrong; the surface is actually a cone because the constant becomes zero after translation.

D.Student A is correct but misidentified the type; it should be a hyperboloid of two sheets.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This comparative reasoning task requires executing and validating algebraic manipulation. Completing the square systematically eliminates linear terms, revealing the canonical form. The resulting equation has two positive and one negative squared term equaling a positive constant, definitively identifying a hyperboloid of one sheet. Student B’s heuristic about mixed signs ignores the role of the constant term.

Q10. When computing the volume enclosed by the ellipsoid $\frac{x^2}{9} + \frac{y^2}{4} + \frac{z^2}{1} = 1$ using triple integrals, a student uses spherical coordinates directly without scaling. Why does this approach fail, and what correction is needed?

A.Spherical coordinates assume radial symmetry; the ellipsoid’s anisotropy requires generalized spherical coordinates with scale factors

a,b,c

in the parametrization. ✅

B.The limits of integration become functions of angles, making evaluation impossible without numerical methods.

C.Spherical coordinates work but require dividing the result by

abc

post-integration to account for stretching.

D.The method works perfectly; the student likely made an arithmetic error in setting up bounds.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Direct spherical coordinates impose $x=\rho\sin\phi\cos\theta$ etc., which only satisfies $x^2+y^2+z^2=\text{const}$ . For ellipsoids, the parametrization must incorporate semi-axes: $x=a\rho\sin\phi\cos\theta$ , etc. This stretches the unit sphere into the ellipsoid, with Jacobian $abc\rho^2\sin\phi$ . Ignoring scaling distorts the domain and yields incorrect volume, highlighting the need for adapted coordinates.

Q11. A quadric surface has circular cross-sections in planes parallel to $z=0$ and also in planes parallel to $x=y$ . Given it passes through the origin and opens upward, which surface must it be?

A.Circular cone

B.Sphere

C.Elliptic paraboloid with equal semi-axes ✅

D.Hyperboloid of revolution

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Circular sections in horizontal planes suggest rotational symmetry about z-axis, implying $x^2 + y^2$ dependence. Circular sections in oblique planes like $x=y$ further constrain the surface: only surfaces of revolution satisfy this for multiple non-parallel families. An upward-opening surface with this property and passing through origin must be a circular paraboloid, i.e., elliptic paraboloid with equal axes.

Q12. In optimizing material usage for a pressurized tank, engineers compare a prolate spheroid and a cylinder with hemispherical ends, both having identical volume and length-to-diameter ratio. Beyond stress distribution, which geometric property of the spheroid makes it superior for minimizing surface area?

A.Its Gaussian curvature is constant everywhere, unlike the cylinder’s zero curvature.

B.Among all surfaces of revolution with fixed volume, the sphere minimizes area; the spheroid approximates this better than composite shapes with discontinuous curvature. ✅

C.The spheroid has fewer edges, reducing manufacturing complexity rather than intrinsic area.

D.Its mean curvature varies smoothly, allowing uniform wall thickness under internal pressure.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This modeling question connects differential geometry to engineering optimization. While cylinders with end caps have piecewise-defined curvature causing stress concentrations, the spheroid’s smooth curvature distribution allows more efficient area-to-volume ratios. The key insight is that deviation from sphericity increases area, but among practical elongated shapes, spheroids outperform composites due to absence of junctions and closer adherence to the isoperimetric principle.

Q13. A student graphs $z = xy$ and insists it cannot be a quadric surface because it lacks squared terms. How would you refute this using coordinate transformation?

A.Rotate coordinates by

45^\circ

in the xy-plane: let

u=(x+y)/\sqrt{2}, v=(x-y)/\sqrt{2}

; then

z = (u^2 - v^2)/2

, revealing a hyperbolic paraboloid. ✅

B.The surface is not quadric; it is a bilinear patch outside the quadric family.

C.Complete the square on

xy

directly to get

(x+y)^2/4 - (x-y)^2/4

, which still lacks pure squares.

D.Multiply both sides by

z

to obtain

z^2 = xyz

, introducing squares artificially.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This addresses a fundamental misconception about canonical forms. Quadrics include cross terms; diagonalization via rotation eliminates them. The $45^\circ$ rotation decouples $xy$ into difference of squares, proving $z=xy$ is indeed a hyperbolic paraboloid. Students must understand that coordinate choice affects appearance, not intrinsic classification, and that cross terms signify rotated principal axes.

Q14. When classifying the surface $2x^2 + 3y^2 + 6z^2 - 4xy - 4xz + 2yz = 1$ , eigenvalue analysis yields one negative and two positive eigenvalues. Without finding eigenvectors, what can be definitively concluded about the surface’s geometry?

A.It is a hyperboloid of two sheets aligned with the eigenvector corresponding to the negative eigenvalue.

B.It is a hyperboloid of one sheet, with the axis of symmetry along the direction of the negative eigenvalue. ✅

C.It is an ellipsoid because all coefficients in the original equation are positive.

D.Classification is impossible without explicit diagonalization.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Eigenvalue signs determine quadric type regardless of orientation. Two positive and one negative eigenvalue with positive constant indicates a hyperboloid of one sheet. The negative eigenvalue’s eigenvector defines the axis along which the surface extends infinitely (the ‘waist’ direction). Original coefficient signs are misleading due to cross terms; spectral analysis is necessary and sufficient for topological classification.

Q15. A researcher models terrain elevation as $z = ax^2 + bxy + cy^2$ . Field measurements show the Hessian determinant $ac - b^2/4 < 0$ at a critical point. What does this imply about local water drainage patterns near that point?

A.Water collects in a basin, indicating a local minimum.

B.Water drains radially outward in all directions, indicating a local maximum.

C.Water flows toward the point along one direction and away along another, creating a saddle-shaped drainage divide. ✅

D.Drainage pattern cannot be determined without higher-order terms.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The Hessian determinant’s sign classifies critical points: negative implies indefinite quadratic form, hence a saddle. In hydrology, saddles act as divides where flow converges from one axis and diverges along the orthogonal axis. This application links multivariable calculus to geomorphology, requiring interpretation of mathematical conditions in physical contexts beyond mere classification labels.

Q16. Which statement correctly distinguishes the asymptotic behavior of $x^2 + y^2 - z^2 = 1$ and $x^2 + y^2 - z^2 = -1$ as $|z| \to \infty$ ?

A.Both approach the same cone

x^2 + y^2 = z^2

, but the first has circular cross-sections growing linearly with |z|, while the second has no real points for |z| < 1. ✅

B.The first surface is bounded in z, while the second extends infinitely.

C.Only the first approaches a cone; the second approaches a cylinder.

D.Their asymptotes differ by a phase shift in the radial growth rate.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Asymptotic cones describe limiting behavior. Both hyperboloids share $x^2+y^2-z^2=0$ as asymptote. However, $=1$ exists for all z with radius $\sqrt{z^2+1}$ , while $=-1$ requires $|z| \geq 1$ , creating a gap. Graphically, one is connected through origin’s neighborhood, the other has two separate lobes. This distinction is vital for understanding global structure from local equations.

Q17. In computer graphics, ray-tracing a quadric involves solving a quadratic equation. If the discriminant is exactly zero for a given ray, what geometric configuration does this represent, and how should rendering handle it?

A.The ray misses the surface entirely; discard the intersection.

B.The ray is tangent to the surface; record a single intersection point for silhouette edge detection. ✅

C.The ray passes through a singularity; apply special smoothing algorithms.

D.Numerical instability occurred; perturb the ray slightly and retry.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Zero discriminant indicates tangency, where the ray grazes the surface. In rendering, this corresponds to silhouette boundaries crucial for outline effects and accurate shading transitions. Misclassifying as miss causes visual artifacts; treating as double intersection wastes computation. Understanding this edge case ensures robust intersection algorithms and highlights the geometric meaning of algebraic conditions in applied contexts.

Q18. A student argues that since $x^2/a^2 + y^2/b^2 = z^2/c^2$ and $x^2/a^2 + y^2/b^2 - z^2/c^2 = 0$ are algebraically identical, their geometric interpretations are interchangeable in all contexts. Why is this problematic in applied settings involving limits or perturbations?

A.They are identical; the student is correct.

B.While algebraically equivalent, the homogeneous form emphasizes conical structure and scaling invariance, whereas the implicit form may obscure singularity at origin in numerical solvers. ✅

C.The first form is only valid for

z \neq 0

, excluding the vertex.

D.The second form cannot be solved explicitly for z, limiting analytical tractability.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Algebraic identity doesn’t guarantee contextual equivalence. The cone’s vertex is a singular point where gradients vanish; numerical methods may fail there despite algebraic validity. Homogeneous representation clarifies projective properties and self-similarity, while implicit form may hide degeneracies. In perturbation or limit analyses, recognizing structural nuances prevents erroneous conclusions about continuity or differentiability at critical locations.

Q19. When fitting a quadric to noisy 3D scan data, least-squares minimization often yields a hyperboloid even when the true object is an ellipsoid. What inherent property of quadric fitting causes this bias, and how might regularization mitigate it?

A.Noise amplifies high-frequency components, favoring unbounded surfaces; adding a penalty on eigenvalue negativity enforces definiteness. ✅

B.The fitting algorithm assumes centered data; translating to centroid eliminates bias.

C.Hyperboloids have more degrees of freedom, so they fit noise better; reducing parameters forces ellipsoidal solutions.

D.The issue stems from poor initialization; using PCA-aligned guesses resolves it.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Unconstrained quadric fitting is ill-conditioned; noise can push eigenvalues across zero, changing surface type. Ellipsoids require all-positive eigenvalues, a non-convex constraint. Regularization penalizing negative eigenvalues or enforcing positive definiteness stabilizes the solution toward physically plausible bounded shapes. This advanced topic bridges numerical analysis and geometric modeling, illustrating why naive algebraic fitting fails in practice.

Q20. Consider the surface defined implicitly by $F(x,y,z) = x^2 + y^2 - z^2 - 2z = 0$ . Without completing the square, how can one determine the location of the narrowest part (waist) of this hyperboloid using calculus?

A.Set partial derivatives to zero; the critical point gives the waist center.

B.Find where the gradient magnitude is minimized along the surface.

C.Solve

\nabla F = \lambda \nabla z

to find extremal z-values on level sets.

D.The waist occurs where

\partial F/\partial z = 0

, yielding

z = -1

; substitute back to find radius. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The waist corresponds to minimal cross-sectional radius. Since cross-sections are circles $x^2+y^2 = z^2+2z$ , minimize $r^2(z) = z^2+2z$ . Derivative $2z+2=0$ gives $z=-1$ . This avoids full completion of square, using calculus on the radial function derived from implicit equation. It demonstrates alternative analytical pathways and reinforces connection between geometry and optimization.

Q21. A physics problem involves equipotential surfaces of a quadrupole field given by $V = k(2z^2 - x^2 - y^2)/r^5$ . Near the origin, the dominant term suggests a quadric approximation. Which surface best approximates the zero-potential locus close to the origin, and why is this approximation valid only locally?

A.A cone, because higher-order terms in

1/r^5

dominate far from origin, distorting the quadric shape. ✅

B.An ellipsoid, since potential must be bounded near sources.

C.A hyperboloid of two sheets, as the sign pattern matches the numerator.

D.The approximation fails entirely because

r^5

introduces essential singularity.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Near origin, $r \to 0$ , but the numerator’s quadratic form dominates angular dependence. Setting $2z^2 - x^2 - y^2 = 0$ gives a cone. However, $1/r^5$ blows up, so the quadric approximation holds only in angular sense, not radially. Farther out, decay alters level set topology. This illustrates asymptotic analysis and limits of local geometric models in singular fields.

Q22. In architectural acoustics, whispering galleries exploit focal properties of ellipsoids. If a room is shaped as a prolate spheroid with foci at $(\pm c, 0, 0)$ , and sound originates at one focus, where does it concentrate, and what happens if the source is displaced slightly off-focus?

A.Sound concentrates precisely at the other focus; displacement causes diffuse scattering with no secondary focus.

B.Sound still focuses sharply at the second focus due to robustness of elliptical geometry.

C.Displacement creates a caustic envelope near the second focus, reducing intensity but maintaining directional concentration. ✅

D.Energy distributes uniformly over the surface, eliminating focal effects.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Perfect focusing occurs only at exact foci. Small displacements generate caustics—envelopes of reflected rays—where energy concentrates in curves rather than points. This degrades acoustic clarity but maintains some directionality. Understanding this sensitivity is crucial for designing functional spaces, linking precise geometry to perceptual outcomes and demonstrating that ideal mathematical properties degrade gracefully under real-world imperfections.

Q23. A student computes the Gaussian curvature of $z = x^2 + y^2$ at the origin and obtains zero, concluding it is flat like a plane. Why is this conclusion incorrect despite the calculation being arithmetically correct?

A.Gaussian curvature at origin is actually 4, not zero; the student used wrong formula.

B.Zero Gaussian curvature implies developability, but the paraboloid is not developable; the student misapplied the definition.

C.At the origin, the surface is umbilic with positive curvature; the student likely computed mean curvature instead.

D.The calculation is correct, but Gaussian curvature alone doesn’t capture extrinsic bending; the surface is intrinsically curved despite K=0 at a point. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Gaussian curvature $K = (f_{xx}f_{yy}-f_{xy}^2)/(1+f_x^2+f_y^2)^2$ . At origin, $f_x=f_y=0$ , $f_{xx}=f_{yy}=2$ , so $K=4$ , not zero. But assuming the student got zero, the deeper issue is confusing pointwise K with global flatness. Even if K=0 at a point, the surface isn’t flat; flatness requires K≡0 everywhere. This highlights distinction between local invariants and global geometry.

Q24. When analyzing the intersection curve of $x^2 + y^2 = 4$ and $z = x^2 - y^2$ , a parameterization $x=2\cos t, y=2\sin t, z=4\cos 2t$ is proposed. What feature of this space curve reveals the underlying quadric’s saddle nature?

A.The z-component oscillates twice as fast as x,y, indicating periodicity unrelated to saddle geometry.

B.The curve lies entirely above z=0, contradicting saddle behavior.

C.The curve achieves both positive and negative z-values symmetrically, reflecting the sign change in

x^2 - y^2

across quadrants. ✅

D.The curve is planar, which cannot occur on a saddle surface.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Substituting the circle into the hyperbolic paraboloid yields $z=4(\cos^2 t - \sin^2 t)=4\cos 2t$ , which ranges from -4 to 4. This sign variation directly mirrors the saddle’s defining trait: positive in some directions, negative in others. Observing this in the intersection curve provides tangible evidence of the surface’s indefinite quadratic form, linking parametric representation to intrinsic geometry.

Q25. In celestial mechanics, orbits are conic sections, but perturbed trajectories may lie on quadric surfaces. If a spacecraft’s trajectory satisfies $x^2 + y^2 = e^2(z + d)^2$ with $e > 1$ , what type of motion does this represent, and how does it differ from Keplerian orbits?

A.It describes a hyperbolic path confined to a cone, representing unbounded escape trajectories with constant eccentricity. ✅

B.It is an elliptical orbit tilted relative to the ecliptic plane.

C.The surface is a paraboloid, indicating marginal escape velocity.

D.This equation defines a cylinder, implying periodic motion unaffected by central force.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The equation $x^2+y^2=e^2(z+d)^2$ with $e>1$ is a cone. Trajectories on cones with constant opening angle correspond to hyperbolic orbits in central force fields, where eccentricity equals the cone’s slope parameter. Unlike Keplerian ellipses (closed), these are open paths. This connects quadric geometry to orbital dynamics, showing how surface classification informs physical interpretation of motion.

Q26. A machine learning model uses quadric decision boundaries in 3D feature space. During training, the learned boundary is $x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = 1$ . Eigenanalysis reveals all eigenvalues are positive. What does this imply about class separability and boundary shape?

A.The boundary is an ellipsoid, indicating compact, well-separated classes with finite decision region. ✅

B.Positive eigenvalues suggest a hyperboloid, implying complex, non-compact separation.

C.The cross terms make classification unreliable regardless of eigenvalues.

D.The surface is degenerate, collapsing to a point, indicating overfitting.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Symmetric matrix with all positive eigenvalues and positive constant defines an ellipsoid. In classification, this means the decision boundary encloses a bounded region, suitable for separating clustered data. Cross terms indicate rotated axes but don’t alter topology. Recognizing this ensures appropriate model interpretation: ellipsoidal boundaries imply convex, finite support regions, contrasting with unbounded hyperboloids that may indicate poor generalization.

Q27. When deriving the surface area of a hyperboloid of one sheet $x^2 + y^2 - z^2 = 1$ between $z = -h$ and $z = h$ , a student uses the formula for surfaces of revolution but forgets the $\sqrt{1+(dz/dr)^2}$ factor. How does this omission quantitatively affect the result?

A.The computed area is too small because the slant height element exceeds the vertical projection. ✅

B.The area is overestimated since the surface curves inward.

C.The error cancels out due to symmetry about z=0.

D.The formula still works because

dz/dr

is constant for hyperboloids.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Surface area of revolution requires arc length element $ds = \sqrt{1+(dr/dz)^2}dz$ or equivalent. Omitting the square root factor replaces true surface element with projected area, systematically underestimating true area. For hyperboloids, $dr/dz = z/r$ , which grows with |z|, so error worsens at extremes. This reinforces that geometric measures demand proper metric tensors, not naive projections.

Q28. In crystallography, optical indicatrices are ellipsoids representing refractive index variation. If a biaxial crystal has principal indices $n_x < n_y < n_z$ , how many circular sections exist in its indicatrix, and what determines their orientation?

A.Infinitely many, as all central sections of ellipsoids are elliptical except special cases.

B.Exactly two, lying in planes containing the intermediate axis

n_y

and bisecting the

n_x

-

n_z

plane. ✅

C.None, because unequal axes prevent circular cross-sections.

D.Exactly one, perpendicular to the optic axis defined by

n_y

.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Biaxial indicatrices have precisely two circular sections, each containing the intermediate axis. These occur where the ellipse formed by intersecting with a plane through $n_y$ has equal semi-axes, requiring specific tilt angles dependent on $n_x,n_z$ . This non-intuitive result arises from quadric geometry: only at particular orientations do principal curvatures match. Knowledge of this is essential for interpreting interference figures in mineralogy.

📝 Quadric surfaces ellipsoid hyperboloid paraboloid (28 MCQs)

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