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πŸ“ Translated quadric surfaces (26 MCQs)

πŸ“– From Calculus β€’ 12. Three Dimensional Space: Vectors β€’ 26 questions available

What is Translated quadric surfaces?

Definition:
Translation replaces x,y,zx,y,z with (xβˆ’h),(yβˆ’k),(zβˆ’l)(x-h),(y-k),(z-l) to shift center/vertex to (h,k,l)(h,k,l), preserving shape but changing position, e.g., (xβˆ’h)2a2+β‹―=1\frac{(x-h)^2}{a^2} + \cdots = 1.

Example:
Sphere (xβˆ’2)2+(y+1)2+(zβˆ’3)2=9(x-2)^2 + (y+1)^2 + (z-3)^2 = 9 is unit sphere scaled by 3 and centered at (2,βˆ’1,3)(2,-1,3) instead of origin.

Reason:
Real-world objects rarely center at origin; translation adapts standard forms to practical coordinates, essential in CAD, geospatial modeling, and fitting data to theoretical surfaces.

4
Easy
14
Medium
8
Hard

πŸ“ All Translated quadric surfaces MCQs

Q1. A student translates the ellipsoid x24+y2+z2=1\frac{x^2}{4} + y^2 + z^2 = 1 by vector ⟨2,βˆ’1,3⟩\langle 2, -1, 3 \rangle but writes (x+2)24+(yβˆ’1)2+(z+3)2=1\frac{(x+2)^2}{4} + (y-1)^2 + (z+3)^2 = 1. Which cognitive error best explains this result?

A.Confusing translation direction with coordinate substitution sign convention βœ…
B.Incorrectly applying scaling factors before translation
C.Misidentifying the center of the original surface
D.Applying rotation matrix instead of translation vector
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The student applied the opposite signs in the substitution. Translating by ⟨h,k,l⟩\langle h,k,l \rangle requires replacing xx with xβˆ’hx-h, not x+hx+h. This is a common sign-reversal misconception when students memorize formulas without understanding that the new coordinates must satisfy the original equation relative to the shifted origin.

Q2. If the hyperboloid x2+y2βˆ’z2=1x^2 + y^2 - z^2 = 1 is translated so its center moves to (1,2,βˆ’3)(1,2,-3), which geometric property remains invariant under this rigid transformation?

A.The orientation of the asymptotic cone
B.The location of the vertices
C.The distance from center to foci along the axis
D.Both A and C βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: Translation preserves all intrinsic metric properties including shape, size, and orientation. The asymptotic cone’s orientation and focal distances are invariant because they depend only on coefficients, not position. Vertices change location but maintain relative positioning. Understanding invariance distinguishes rigid motions from deformations in quadric surface analysis.

Q3. An engineer models a cooling tower as (xβˆ’5)29+(z+2)29βˆ’(yβˆ’4)216=1\frac{(x-5)^2}{9} + \frac{(z+2)^2}{9} - \frac{(y-4)^2}{16} = 1. To reposition it so the narrowest cross-section lies in the plane y=0y=0, what translation vector is required?

A.⟨0,βˆ’4,0⟩\langle 0, -4, 0 \rangle
B.⟨0,4,0⟩\langle 0, 4, 0 \rangle βœ…
C.βŸ¨βˆ’5,0,2⟩\langle -5, 0, 2 \rangle
D.⟨5,βˆ’4,βˆ’2⟩\langle 5, -4, -2 \rangle
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: The narrowest cross-section of a hyperboloid of one sheet occurs at its center where y=4y=4. Moving this to y=0y=0 requires subtracting 4 from y-coordinates, meaning we add 4 to the variable: replace yy with yβˆ’4y-4 becomes yy, requiring translation vector ⟨0,4,0⟩\langle 0,4,0 \rangle. This tests application of geometric features to real-world repositioning.

Q4. Given two paraboloids P1:z=x2+y2P_1: z = x^2 + y^2 and P2:zβˆ’3=(xβˆ’1)2+(y+2)2P_2: z-3 = (x-1)^2 + (y+2)^2, which statement correctly compares their geometric relationship?

A.P2 is P1 translated by ⟨1,βˆ’2,3⟩\langle 1,-2,3 \rangle with identical curvature βœ…
B.P2 is P1 translated and vertically stretched
C.P2 has different focal length due to translation
D.P1 and P2 intersect in exactly one point
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Comparing equations shows P2 results from substituting xβ†’xβˆ’1x \to x-1, yβ†’y+2y \to y+2, zβ†’zβˆ’3z \to z-3 in P1, confirming pure translation by ⟨1,βˆ’2,3⟩\langle 1,-2,3 \rangle. Translation preserves all differential properties including Gaussian curvature and focal parameters. Students often mistakenly believe position affects shape; this question targets that conceptual gap through direct comparison.

Q5. A quadric surface has equation 4(xβˆ’1)2+9(y+2)2+z2=364(x-1)^2 + 9(y+2)^2 + z^2 = 36. After translating to standard position centered at origin, what is the length of the semi-axis along the y-direction?

A.2 βœ…
B.3
C.4
D.6
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: Dividing by 36 gives (xβˆ’1)29+(y+2)24+z236=1\frac{(x-1)^2}{9} + \frac{(y+2)^2}{4} + \frac{z^2}{36} = 1. Translation doesn’t alter axis lengths; the y-semi-axis is 4=2\sqrt{4} = 2. This direct recall question verifies foundational knowledge that translation affects only position terms, not denominators representing squared semi-axes. Essential baseline for higher-order problems involving transformed quadrics.

Q6. When analyzing the surface (xβˆ’a)2A+(yβˆ’b)2Bβˆ’(zβˆ’c)2C=1\frac{(x-a)^2}{A} + \frac{(y-b)^2}{B} - \frac{(z-c)^2}{C} = 1, a student claims changing a,b,ca,b,c alters the surface type. Which rebuttal most accurately addresses this misconception?

A.Surface type depends solely on signs and number of positive/negative terms
B.Translation parameters only shift the center, preserving classification
C.Both A and B are correct and equivalent statements βœ…
D.The claim is actually true for hyperboloids but false for ellipsoids
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: Quadric classification is determined by the signature of the quadratic form (signs of eigenvalues), unaffected by linear translation terms. Changing a,b,ca,b,c merely relocates the center while maintaining the same canonical form. Options A and B express this invariant principle differently; recognizing their equivalence demonstrates deep conceptual understanding beyond rote memorization of classification rules.

Q7. A satellite dish is modeled by z=18(x2+y2)z = \frac{1}{8}(x^2 + y^2). If installed at height 10m above ground with focus at receiver located at (0,0,12)(0,0,12), what translation was applied to the standard paraboloid z=18(x2+y2)z = \frac{1}{8}(x^2 + y^2)?

A.⟨0,0,10⟩\langle 0,0,10 \rangle βœ…
B.⟨0,0,2⟩\langle 0,0,2 \rangle
C.⟨0,0,8⟩\langle 0,0,8 \rangle
D.No translation; focus adjustment changes parameter
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Standard paraboloid z=14p(x2+y2)z = \frac{1}{4p}(x^2+y^2) has focus at (0,0,p)(0,0,p). Here p=2p=2, so unshifted focus is at z=2z=2. Receiver at z=12z=12 implies vertical shift of 10 units. Thus translation vector is ⟨0,0,10⟩\langle 0,0,10 \rangle. This multi-step problem combines focal property knowledge with translation application in engineering context, testing integrated modeling skills.

Q8. Consider the graph of an elliptic paraboloid opening upward with vertex at (3,βˆ’1,5)(3,-1,5). If the contour lines in horizontal planes appear as concentric ellipses centered at x=3,y=βˆ’1x=3, y=-1, what can be definitively concluded about the translation?

A.Only the z-coordinate of translation affects contour spacing
B.The xy-translation determines contour center; z-translation sets vertex height βœ…
C.Contour shape reveals both translation and scaling factors
D.Horizontal contours cannot determine any translation component
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Contour centers directly reflect xy-translation since level curves z=kz=k yield (xβˆ’h)2a2+(yβˆ’k)2b2=kβˆ’c\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = k-c, centered at (h,k)(h,k). The z-value where contours begin indicates vertical translation. This graph-based interpretation connects visual pattern recognition to algebraic parameters, assessing ability to extract translation information from projected views rather than explicit equations.

Q9. Two students derive translations for moving x2+2y2+3z2=6x^2 + 2y^2 + 3z^2 = 6 to center at (1,βˆ’2,4)(1,-2,4). Student A writes (xβˆ’1)2+2(y+2)2+3(zβˆ’4)2=6(x-1)^2 + 2(y+2)^2 + 3(z-4)^2 = 6; Student B expands then completes square. Why is Student A’s method superior for verification?

A.Avoids arithmetic errors in expansion/completion steps
B.Preserves immediate visibility of translation parameters
C.Both reasons are valid and complementary βœ…
D.Student B’s method is actually more reliable
πŸ’‘ Difficulty: easy | βœ… Correct: C

πŸ“– Explanation: Direct substitution maintains structural clarity, allowing instant validation that each variable shift matches intended translation. Expansion risks computational mistakes and obscures parameter correspondence. However, completing squares serves as independent verification. Recognizing complementary strengths of methods reflects metacognitive awareness of mathematical practice, going beyond procedural execution to strategic tool selection based on purpose.

Q10. A hyperbolic paraboloid z=xyz = xy is translated to zβˆ’2=(xβˆ’1)(y+3)z-2 = (x-1)(y+3). What is the new location of the saddle point?

A.(1,βˆ’3,2)(1,-3,2) βœ…
B.(1,3,2)(1,3,2)
C.(βˆ’1,3,βˆ’2)(-1,3,-2)
D.(1,βˆ’3,βˆ’2)(1,-3,-2)
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: The saddle point of z=xyz=xy is at origin. Substituting X=xβˆ’1X=x-1, Y=y+3Y=y+3, Z=zβˆ’2Z=z-2 transforms to Z=XYZ=XY, whose saddle is at X=Y=Z=0X=Y=Z=0. Solving gives x=1,y=βˆ’3,z=2x=1, y=-3, z=2. This requires understanding that saddle points translate like centers despite lacking symmetry axes, testing nuanced grasp of critical point behavior under affine transformations.

Q11. In optimizing antenna placement, engineers use (xβˆ’10)225+(y+5)216+(zβˆ’8)29=1\frac{(x-10)^2}{25} + \frac{(y+5)^2}{16} + \frac{(z-8)^2}{9} = 1. If safety regulations require minimum 3-unit clearance from building face at x=0x=0, does current positioning comply?

A.Yes; closest point is at x=5 > 3 βœ…
B.No; closest point is at x=5 but clearance measured from surface not center
C.Yes; center at x=10 provides adequate buffer
D.Insufficient information without knowing ellipsoid orientation
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Ellipsoid extends Β±5 in x-direction from center x=10, so minimum x-coordinate is 10βˆ’5=5. Clearance from x=0 plane is 5 units, exceeding 3-unit requirement. This scenario-based problem integrates geometric extent calculation with regulatory constraint checking, demanding careful distinction between center position and surface boundary in applied contexts. Misconception about measuring from center versus surface is targeted in distractors.

Q12. Which transformation sequence converts 4x2+y2+9z2=364x^2 + y^2 + 9z^2 = 36 into X29+Y236+Z24=1\frac{X^2}{9} + \frac{Y^2}{36} + \frac{Z^2}{4} = 1 with center at (2,βˆ’1,3)(2,-1,3)?

A.Scale then translate βœ…
B.Translate then scale
C.Simultaneous affine transformation
D.Impossible; axis ratios don’t match
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Original normalized form is x29+y236+z24=1\frac{x^2}{9} + \frac{y^2}{36} + \frac{z^2}{4} = 1. Target has same denominators but permuted variables and shifted center. Scaling isn't needed since semi-axes already match target values after normalization. Only translation by ⟨2,βˆ’1,3⟩\langle 2,-1,3 \rangle and relabeling axes achieves result. However, if interpreted as literal variable mapping, scaling precedes translation in standard reduction procedure. This mixed-concept question tests understanding of canonical form derivation order.

Q13. A student argues that translating z=x2βˆ’y2z = x^2 - y^2 by ⟨a,b,c⟩\langle a,b,c \rangle changes its Gaussian curvature at corresponding points. Which response correctly evaluates this claim?

A.True; curvature depends on absolute position in space
B.False; Gaussian curvature is invariant under rigid motions βœ…
C.Partially true; mean curvature changes but Gaussian doesn’t
D.Depends on whether translation includes z-component
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Gaussian curvature is an intrinsic property preserved under isometries including translations. For z=x2βˆ’y2z=x^2-y^2, K=βˆ’4(1+4x2+4y2)2K = \frac{-4}{(1+4x^2+4y^2)^2}; after translation, expression shifts arguments but functional form and values at corresponding points remain identical. This challenges advanced misconception linking extrinsic position to intrinsic geometry, requiring differential geometry insight beyond basic quadric classification.

Q14. Given contour plot showing elliptical level curves centered at (2,βˆ’3)(2,-3) with increasing density toward center, and knowing surface is quadric, what additional information is needed to uniquely determine the translation vector?

A.The z-value at the center point βœ…
B.Orientation of major/minor axes
C.Whether surface opens up or down
D.All of the above
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Contour center gives xy-translation (2,βˆ’3)(2,-3). Density gradient indicates extremum type but not absolute z-position. Without knowing zz-coordinate at center, vertical translation remains undetermined. Axis orientation and opening direction affect shape classification but not translation parameters once center xy is known. This graph-based inference question highlights insufficiency of partial visual data and necessity of combining multiple cues for complete parameter recovery.

Q15. During error analysis of lab data fitting, residuals show systematic bias when using x2a2+y2b2+z2c2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 instead of translated form. What diagnostic step most efficiently identifies missing translation?

A.Check if residual mean is nonzero across dataset
B.Examine residual correlation with coordinates βœ…
C.Plot residuals against predicted values
D.Increase polynomial degree in fit
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Nonzero residual mean suggests constant offset, but correlation with x,y,zx,y,z specifically reveals linear trends indicating unmodeled translation. If residuals increase with xx, likely xx-translation missing. This targeted diagnostic leverages statistical patterns to pinpoint specific parameter omission, demonstrating applied error analysis skill beyond generic goodness-of-fit assessment. Correlation structure encodes directional information absent in aggregate statistics.

Q16. An architect designs a dome using upper half of (xβˆ’4)216+(y+3)29+(zβˆ’10)225=1\frac{(x-4)^2}{16} + \frac{(y+3)^2}{9} + \frac{(z-10)^2}{25} = 1. If foundation requires base circle at z=0z=0, what horizontal cross-section radius exists at ground level?

A.No intersection; dome floats above ground
B.Ellipse with semi-axes 3.2 and 1.8 βœ…
C.Circle of radius approximately 2.4
D.Cannot determine without full ellipsoid
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Set z=0z=0: (xβˆ’4)216+(y+3)29=1βˆ’10025=βˆ’3\frac{(x-4)^2}{16} + \frac{(y+3)^2}{9} = 1 - \frac{100}{25} = -3. Negative right side means no real solution; dome minimum z is 10βˆ’5=510-5=5. But waitβ€”recalculating: (zβˆ’10)225≀1\frac{(z-10)^2}{25} \leq 1 implies z∈[5,15]z \in [5,15]. At z=0z=0, LHS >1 always. Actually, no intersection exists. Correction: Option A is correct. This trick question tests careful domain analysis before computation, exposing tendency to mechanically solve without checking feasibilityβ€”a critical HOTS in modeling validation.

Q17. Which statement best synthesizes the role of translation in quadric surface taxonomy?

A.Translation generates distinct surfaces within same class
B.Translation produces congruent surfaces differing only in position βœ…
C.Translation can convert one quadric type to another
D.Translation is irrelevant to classification but essential for applications
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: All quadrics related by translation are geometrically congruent; classification depends exclusively on quadratic form signature. Translation creates positional variants within equivalence classes defined by invariants. This conceptual synthesis integrates algebraic, geometric, and categorical perspectives, forming foundational understanding for advanced topics like moduli spaces. Distractors represent common overgeneralizations or underappreciations of translation’s limited but precise role.

Q18. In computer graphics, rendering efficiency improves when quadrics are centered at origin. Given 2(xβˆ’3)2+3(y+1)2+(zβˆ’5)2=122(x-3)^2 + 3(y+1)^2 + (z-5)^2 = 12, what preprocessing translation minimizes shader complexity?

A.βŸ¨βˆ’3,1,βˆ’5⟩\langle -3,1,-5 \rangle βœ…
B.⟨3,βˆ’1,5⟩\langle 3,-1,5 \rangle
C.⟨0,0,0⟩\langle 0,0,0 \rangle; no translation needed
D.Normalize coefficients first, then translate
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: To center at origin, apply inverse translation: replace xβ†’x+3x \to x+3, etc., corresponding to moving surface by βŸ¨βˆ’3,1,βˆ’5⟩\langle -3,1,-5 \rangle. This eliminates linear terms in shader code. Option D misorders operations; normalization doesn’t affect center. Graphics optimization context makes abstract translation concrete, testing transfer of mathematical concept to computational implementation where performance depends on canonical positioning.

Q19. A physics problem involves potential V=(xβˆ’2)2+4(y+1)2+9(zβˆ’3)2V = (x-2)^2 + 4(y+1)^2 + 9(z-3)^2. Equilibrium occurs where gradient vanishes. How does translation simplify finding equilibrium compared to expanding?

A.Eliminates need for calculus entirely
B.Reduces system to trivial solution at translated origin
C.Makes Hessian diagonal automatically
D.Both B and C βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: Gradient βˆ‡V=⟨2(xβˆ’2),8(y+1),18(zβˆ’3)⟩\nabla V = \langle 2(x-2), 8(y+1), 18(z-3) \rangle vanishes immediately at (2,βˆ’1,3)(2,-1,3). Expansion yields coupled linear system requiring solving. Translation reveals equilibrium as center by construction. Hessian is already diagonal regardless, but translation makes critical point obvious. This mixed-concept question links multivariable calculus to geometric insight, showing how structural recognition supersedes brute-force computation in identifying physical equilibria.

Q20. Olympiad Challenge: Find all vectors vβƒ—\vec{v} such that translating x2+y2βˆ’z2=1x^2 + y^2 - z^2 = 1 by vβƒ—\vec{v} yields a surface containing the line x=t,y=t,z=tx=t, y=t, z=t. Express answer parametrically.

A.vβƒ—=⟨a,a,a⟩\vec{v} = \langle a,a,a \rangle for any a∈Ra \in \mathbb{R}
B.vβƒ—=⟨a,b,c⟩\vec{v} = \langle a,b,c \rangle where a2+b2βˆ’c2=1a^2 + b^2 - c^2 = 1
C.vβƒ—=⟨a,a,Β±2a2βˆ’1⟩\vec{v} = \langle a,a,\pm\sqrt{2a^2-1} \rangle for ∣a∣β‰₯1/2|a| \geq 1/\sqrt{2} βœ…
D.No such translation exists
πŸ’‘ Difficulty: hard | βœ… Correct: C

πŸ“– Explanation: Translated surface: (xβˆ’a)2+(yβˆ’b)2βˆ’(zβˆ’c)2=1(x-a)^2 + (y-b)^2 - (z-c)^2 = 1. Line satisfies iff (tβˆ’a)2+(tβˆ’b)2βˆ’(tβˆ’c)2=1(t-a)^2 + (t-b)^2 - (t-c)^2 = 1 for all t. Expanding: t2(1+1βˆ’1)+t(βˆ’2aβˆ’2b+2c)+(a2+b2βˆ’c2βˆ’1)=0t^2(1+1-1) + t(-2a-2b+2c) + (a^2+b^2-c^2-1) = 0. Coefficient of t2t^2 is 1β‰ 0, impossible unless... Wait, recalculate: coefficient is 1, so identity requires 1=0 contradiction. But if line lies on surface, equation must hold identically. Since leading coefficient β‰ 0, no solution exists. However, reconsider: maybe specific t values? No, β€œcontaining the line” means all t. Thus D seems correct. But let’s verify special case: if a=b=c, then expression becomes t2βˆ’2at+(2a2βˆ’a2βˆ’1)=t2βˆ’2at+(a2βˆ’1)t^2 - 2at + (2a^2 - a^2 -1) = t^2 -2at + (a^2-1), still quadratic. Indeed, no translation works because hyperboloid contains no straight lines through arbitrary directions? Actually hyperboloid of one sheet contains two families of lines, but not necessarily this diagonal. Final answer: D. This Olympiad-style problem demands rigorous identity analysis and knowledge of ruled surface properties, pushing beyond standard curriculum.

Q21. When comparing numerical integration accuracy over translated vs. untranslated ellipsoids, why might centered-at-origin version yield better results even with identical mesh density?

A.Symmetric cancellation of truncation errors
B.Reduced condition number of Jacobian
C.Alignment with coordinate grid reduces interpolation error
D.All listed factors contribute βœ…
πŸ’‘ Difficulty: hard | βœ… Correct: D

πŸ“– Explanation: Centered quadrics exploit symmetry for error cancellation in quadrature rules. Coordinate alignment minimizes discretization artifacts when grids are axis-aligned. Well-conditioned parametrizations near origin improve numerical stability. These combined effects make canonical positioning advantageous computationally. This advanced application question bridges theoretical geometry with numerical analysis, illustrating how mathematical structure influences algorithmic performance beyond mere representation convenience.

Q22. A researcher fits experimental data to Ax2+By2+Cz2+Dx+Ey+Fz+G=0Ax^2 + By^2 + Cz^2 + Dx + Ey + Fz + G = 0 and obtains nonzero D,E,F. Before classifying, what essential step must precede type determination?

A.Normalize so largest coefficient is 1
B.Complete squares to eliminate linear terms βœ…
C.Compute discriminant of quadratic form
D.Rotate to principal axes
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Linear terms indicate off-center position; classification requires canonical form achieved by completing squares to isolate translation. Only after removing D,E,F via translation can signature analysis correctly identify surface type. Rotation handles cross-terms, not linear ones. This procedural sequencing question tests understanding that translation reduction is prerequisite to classification, addressing common student error of attempting type identification on non-canonical forms.

Q23. In robotics path planning, obstacle modeled as (xβˆ’5)24+(y+2)29+z21≀1\frac{(x-5)^2}{4} + \frac{(y+2)^2}{9} + \frac{z^2}{1} \leq 1. Robot sphere radius 0.5 needs safe passage. What effective translation describes Minkowski sum boundary?

A.Same center; semi-axes increased by 0.5
B.Center unchanged; equation becomes (xβˆ’5)2(2.5)2+(y+2)2(3.5)2+z2(1.5)2≀1\frac{(x-5)^2}{(2.5)^2} + \frac{(y+2)^2}{(3.5)^2} + \frac{z^2}{(1.5)^2} \leq 1 βœ…
C.Translate obstacle by ⟨0,0,0.5⟩\langle 0,0,0.5 \rangle
D.Cannot represent as single quadric
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Minkowski sum of ellipsoid and sphere yields larger concentric ellipsoid with semi-axes incremented by sphere radius. Translation components unchanged since sphere is symmetric. New semi-axes: 2+0.5=2.52+0.5=2.5, 3+0.5=3.53+0.5=3.5, 1+0.5=1.51+0.5=1.5. This spatial reasoning problem applies translation concepts to collision avoidance, requiring understanding that uniform dilation preserves center while modifying scale parametersβ€”a non-obvious extension of basic translation knowledge to geometric operations.

Q24. Student submits work claiming (xβˆ’1)2+(yβˆ’2)2+(zβˆ’3)2=0(x-1)^2 + (y-2)^2 + (z-3)^2 = 0 represents a sphere of radius 1 centered at (1,2,3). What fundamental misunderstanding does this reveal?

A.Confusing degenerate case with standard sphere βœ…
B.Misapplying translation formula signs
C.Believing zero-radius spheres exist as valid surfaces
D.Thinking RHS determines center not radius
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: Equation equals zero only at single point (1,2,3); it’s degenerate, not a sphere. Standard sphere requires positive RHS equal to radius squared. This error-analysis question targets confusion between limiting cases and proper definitions, emphasizing that translation alone cannot create non-degenerate objects from degenerate ones. Recognizing degeneracy is crucial for avoiding invalid geometric interpretations in proofs and applications.

Q25. Given family of surfaces (xβˆ’t)24+y29+z2=1\frac{(x-t)^2}{4} + \frac{y^2}{9} + z^2 = 1 parameterized by t, describe the locus of centers as t varies.

A.Line parallel to x-axis through (0,0,0) βœ…
B.Plane x=0
C.Single fixed point (0,0,0)
D.Helix around z-axis
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Center is at (t,0,0)(t,0,0) for each t, tracing x-axis. Translation parameter t directly controls center x-coordinate while y,z centers remain fixed. This dynamic perspective question shifts focus from static surfaces to parametric families, testing understanding of translation as continuous deformation generator. Locus characterization connects individual surface properties to collective behavior, fostering higher-level abstraction skills essential for advanced geometry.

Q26. In verifying CAD model integrity, software flags inconsistency when user specifies ellipsoid center at (1,2,3) but equation shows (x+1)24+(yβˆ’2)29+(z+3)216=1\frac{(x+1)^2}{4} + \frac{(y-2)^2}{9} + \frac{(z+3)^2}{16} = 1. What automated check would catch this discrepancy?

A.Compare extracted center from completed-square form against declared center βœ…
B.Verify all denominators are positive
C.Ensure no cross-product terms present
D.Check that trace of quadratic form matches volume
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: Automated validation should parse equation into canonical form via completing squares, extract actual center (βˆ’1,2,βˆ’3)(-1,2,-3), and compare to user-declared (1,2,3). Sign mismatch in linear terms causes center inversion. This quality-control scenario emphasizes programmatic verification of human input, translating mathematical consistency checks into software logic. It reinforces that translation parameters must align precisely between symbolic and metadata representations in technical workflows.

πŸ”— Related Topics (MCQs)