CELE Engineering Mathematics — Analytic GeometryExam Answer Templates

Examiner tip

At 1-mark level, only the exact correct formula earns the mark. There is no partial credit — write it completely and correctly in one line.

Model answer

d = √[(x₂ − x₁)² + (y₂ − y₁)²]

Answer structure

• Line 1: Write the distance formula correctly with radical sign enclosing both squared difference terms. [1 mark]

Scoring breakdown

Common mark deductions

• Writing d = (x₂−x₁) + (y₂−y₁) — omitting squares and radical (−1 mark, no credit).
• Writing d² instead of d (formula for d², not d — half-credit at best).
• Omitting the radical sign.

Key phrases to include

• d = √[(x₂−x₁)² + (y₂−y₁)²]
• square root
• sum of squares of differences

Examiner tip

Memorize the eccentricity table: e = 0 (circle), 0 < e < 1 (ellipse), e = 1 (parabola), e > 1 (hyperbola). This is a frequent 1-mark item.

Model answer

For an ellipse, the eccentricity e satisfies 0 < e < 1, and the relationship between the semi-axes and focal distance is c² = a² − b².

Answer structure

• Part 1: State eccentricity range 0 < e < 1. [0.5 mark]
• Part 2: State c² = a² − b². [0.5 mark]

Scoring breakdown

Common mark deductions

• Writing c² = a² + b² (hyperbola formula — classic confusion, −1 mark).
• Stating e < 1 without the lower bound e > 0.
• Confusing a and b (a is always the semi-major axis, a > b).

Key phrases to include

• 0 < e < 1
• c² = a² − b²
• eccentricity

Examiner tip

For 2-mark numericals, the formula mark and the answer mark are separate. Even if you get the arithmetic wrong, you still earn 1 mark for the correct formula.

Model answer

Given: A(x₁, y₁) = (2, 5), B(x₂, y₂) = (6, 13)
Required: Slope m

Solution:
m = (y₂ − y₁) / (x₂ − x₁)
m = (13 − 5) / (6 − 2)
m = 8 / 4

∴ m = 2

Answer structure

• Line 1: Write 'Given:' — identify coordinates of both points. [setup, no mark yet]
• Line 2: Write 'Required:' — state what is to be found.
• Line 3: State the slope formula m = (y₂−y₁)/(x₂−x₁). [1 mark — formula]
• Line 4: Substitute values and simplify. [1 mark — correct substitution and answer]

Scoring breakdown

Common mark deductions

• Inverting the formula: writing (x₂−x₁)/(y₂−y₁) — loses formula mark.
• Arithmetic error: 13−5 = 8, 6−2 = 4, so m = 8/4 = 2 — any arithmetic slip loses the answer mark.
• Not labeling which point is (x₁,y₁) vs (x₂,y₂) — acceptable either order as long as consistent.

Key phrases to include

• m = (y₂−y₁)/(x₂−x₁)
• slope m = 2
• rise over run

Examiner tip

Always verify the sign of C carefully. If the line is given as 3x + 4y = 10, rewrite it as 3x + 4y − 10 = 0 so C = −10, not +10.

Model answer

Given: Point P(x₀, y₀) = (3, 4); Line: 3x + 4y − 10 = 0 → A = 3, B = 4, C = −10
Required: Perpendicular distance d

Solution:

d = 15 / 5

∴ d = 3 units

Answer structure

• Line 1: Identify A, B, C from line equation (must show the line is already in Ax+By+C=0 form). [setup]
• Line 2: Write the point-to-line distance formula. [1 mark]
• Line 3-4: Substitute and simplify, keeping absolute value bars until the final numerical value. [1 mark]

Scoring breakdown

Common mark deductions

• Omitting absolute value bars — sign error can give negative distance (physically impossible).
• Using √(A+B) instead of √(A²+B²) — loses formula mark.
• Not rewriting line in standard Ax+By+C=0 form first — risk of sign error in C.

Key phrases to include

• d = |Ax₀ + By₀ + C| / √(A² + B²)
• absolute value
• d = 3 units

Examiner tip

For 3-mark problems with two parts, split your solution clearly into Part 1 and Part 2. Examiners mark each part independently — a wrong midpoint does not prevent you from earning the distance mark.

Model answer

Given: P₁(1, 2), P₂(7, 10)
Required: Midpoint M and length d

Solution — Part 1 (Midpoint):
M = ((x₁+x₂)/2, (y₁+y₂)/2)
M = ((1+7)/2, (2+10)/2)
M = (8/2, 12/2)
M = (4, 6)

Solution — Part 2 (Distance):
d = √[(x₂−x₁)² + (y₂−y₁)²]
d = √[(7−1)² + (10−2)²]
d = √[36 + 64]
d = √100

∴ Midpoint M = (4, 6) and Length d = 10 units

Answer structure

• Line 1: Write midpoint formula. [1 mark]
• Line 2: Substitute and compute midpoint M = (4, 6). [1 mark]
• Line 3: Write distance formula, substitute, and compute d = 10. [1 mark]

Scoring breakdown

Common mark deductions

• Adding coordinates instead of averaging: (1+7, 2+10) = (8, 12) — not the midpoint.
• Forgetting to square both differences in the distance formula.
• Not simplifying √100 to 10 — leaving the answer as √100 may lose the final answer mark.

Key phrases to include

• M = ((x₁+x₂)/2, (y₁+y₂)/2)
• d = √[(x₂−x₁)² + (y₂−y₁)²]
• midpoint M = (4, 6)
• d = 10 units

Examiner tip

Always clear fractions when writing in general form — multiply through by the LCD. The general form must have integer coefficients with no common factor (unless 1).

Model answer

Given: Points (1, 2) and (4, 6)
Required: Equation of line in general form

Step 1 — Compute slope:
m = (y₂−y₁)/(x₂−x₁) = (6−2)/(4−1) = 4/3

Step 2 — Point-slope form using (1, 2):
y − 2 = (4/3)(x − 1)
3(y − 2) = 4(x − 1) [multiply both sides by 3]
3y − 6 = 4x − 4

Step 3 — Rearrange to general form:
4x − 3y − 4 + 6 = 0

∴ 4x − 3y + 2 = 0

Answer structure

• Step 1: Compute slope m = 4/3. [1 mark]
• Step 2: Apply point-slope form y−y₁ = m(x−x₁) and substitute. [1 mark]
• Step 3: Simplify to Ax + By + C = 0 form. [1 mark]

Scoring breakdown

Common mark deductions

• Leaving answer in slope-intercept form (y = 4x/3 + ...) when general form is specifically requested.
• Sign error when moving terms across the equals sign.
• Not clearing fractions — leaving coefficients as fractions in the general form.

Key phrases to include

• m = (y₂−y₁)/(x₂−x₁)
• y − y₁ = m(x − x₁)
• 4x − 3y + 2 = 0

Examiner tip

The most common error is the sign of k. If center is (h, k) = (2, −3), the equation is (x−2)² + (y−(−3))² = (y+3)². Always expand the sign explicitly.

Model answer

Given: Center C(2, −3), Point on circle P(5, 1)
Required: Radius r and standard equation of circle

Step 1 — Compute radius r = distance from center to point:
r = √[(5−2)² + (1−(−3))²]
r = √[3² + 4²]
r = √[9 + 16]
r = √25 = 5

Step 2 — Write standard form (x−h)² + (y−k)² = r²:
(x−2)² + (y−(−3))² = 5²

∴ (x−2)² + (y+3)² = 25

Answer structure

• Step 1: Compute r using distance formula between center and given point. [1 mark]
• Step 2: Identify h = 2, k = −3, r = 5. [setup]
• Step 3: Write standard circle equation (x−2)² + (y+3)² = 25. [2 marks — 1 for form, 1 for correct substitution]

Scoring breakdown

Common mark deductions

• Writing (y−3)² instead of (y+3)² — sign error with negative k.
• Writing r instead of r² on the right-hand side.
• Forgetting to compute r; just assuming r = distance formula without showing work.

Key phrases to include

• (x−h)² + (y−k)² = r²
• r = 5
• (x−2)² + (y+3)² = 25
• center (2, −3)

Examiner tip

An alternative correct method is to pick any point on L₁ and compute its distance to L₂ using the point-to-line formula — this gives the same answer and is equally acceptable.

Model answer

Given: L₁: 3x + 4y − 10 = 0 and L₂: 3x + 4y + 5 = 0
Required: Distance d between parallel lines

Note: Both lines have the same coefficients A = 3, B = 4, confirming they are parallel.

Step 1 — Formula for distance between parallel lines Ax+By+C₁=0 and Ax+By+C₂=0:
d = |C₁ − C₂| / √(A² + B²)

Step 2 — Identify constants: C₁ = −10, C₂ = +5

Step 3 — Substitute:

d = 15 / 5

∴ d = 3 units

Answer structure

• Step 1: Verify lines are parallel (same A and B coefficients). [0.5 mark]
• Step 2: State formula d = |C₁−C₂| / √(A²+B²). [1 mark]
• Step 3: Substitute C₁ = −10, C₂ = 5, compute d = 3. [1.5 marks]

Scoring breakdown

Common mark deductions

• Using |C₁+C₂| instead of |C₁−C₂|.
• Mixing up the constants: using +10 instead of −10 for C₁.
• Not confirming that lines are parallel before applying the formula.

Key phrases to include

• d = |C₁−C₂| / √(A²+B²)
• parallel lines — same A and B
• d = 3 units
• absolute value

Examiner tip

Always check: whichever denominator is larger, a² is under that variable and the foci lie along that axis. For x²/25 + y²/9: 25 > 9, so major axis is along x-axis, foci are on x-axis.

Model answer

Given: x²/25 + y²/9 = 1
Required: Foci and eccentricity

Step 1 — Identify a² and b²:
Standard form: (x−h)²/a² + (y−k)²/b² = 1, center at origin (0,0)
Since 25 > 9: a² = 25, b² = 9
∴ a = 5, b = 3 (major axis along x-axis)

Step 2 — Find c using ellipse relation:
c² = a² − b² = 25 − 9 = 16
c = 4

Step 3 — Foci (along x-axis since a² under x):
F₁(−4, 0) and F₂(4, 0)

Step 4 — Eccentricity:
e = c/a = 4/5 = 0.80

∴ Foci: F(±4, 0); Eccentricity e = 0.80 (< 1, confirms ellipse)

Answer structure

• Step 1: Identify a² = 25, b² = 9, center at origin. [1 mark]
• Step 2: Compute c = √(a²−b²) = 4. [1 mark]
• Step 3: State foci F(±4, 0) and eccentricity e = 4/5 = 0.80. [1 mark]

Scoring breakdown

Common mark deductions

• Using c² = a² + b² (hyperbola formula) — most critical error.
• Placing foci on the y-axis without checking which denominator is larger.
• Not confirming e < 1 for an ellipse.

Key phrases to include

• c² = a² − b²
• a² = 25, b² = 9
• foci F(±4, 0)
• e = c/a = 4/5 = 0.80

Examiner tip

A 5-mark problem expects all five logical steps written clearly. Do not skip any step even if it seems obvious. Each step is worth a mark. Always verify by back-substituting the midpoint into your final equation.

Model answer

Given: A(2, 1) and B(6, 5)
Required: Equation of the perpendicular bisector of AB

Step 1 — Find midpoint M of AB:
M = ((2+6)/2, (1+5)/2) = (8/2, 6/2) = (4, 3)

Step 2 — Find slope of AB:
m_AB = (5−1)/(6−2) = 4/4 = 1

Step 3 — Find slope of perpendicular bisector:
For perpendicular lines: m₁ × m₂ = −1
m_perp = −1/m_AB = −1/1 = −1

Step 4 — Write equation of perpendicular bisector through M(4,3) with slope −1:
y − 3 = −1(x − 4)
y − 3 = −x + 4
x + y − 7 = 0

Verification: Check that M(4,3) lies on x + y − 7 = 0:
4 + 3 − 7 = 0 ✓

∴ Perpendicular bisector: x + y − 7 = 0

Answer structure

• Step 1: Compute midpoint M = (4, 3). [1 mark]
• Step 2: Compute slope of AB: m_AB = 1. [1 mark]
• Step 3: Apply perpendicular slope condition: m_perp = −1. [1 mark]
• Step 4: Apply point-slope form through M with m_perp, simplify to x + y − 7 = 0. [1 mark]
• Step 5: Verification — substitute M into equation. [1 mark — or awarded for correct final form]

Scoring breakdown

Common mark deductions

• Using negative of slope instead of negative reciprocal — e.g., writing m_perp = −1 when m_AB = 1 is coincidentally correct here, but writing m_perp = −m instead of −1/m is a conceptual error that would fail for other slopes.
• Substituting original point A or B instead of midpoint M into point-slope form.
• Not converting to general form when specifically required.
• Arithmetic error in midpoint — averaging incorrectly.

Key phrases to include

• midpoint M = (4, 3)
• m₁ × m₂ = −1
• perpendicular slope m_perp = −1
• y − y₁ = m(x − x₁)
• x + y − 7 = 0

Examiner tip

For 5-mark conic problems, always state explicitly whether c² = a²−b² (ellipse) or c² = a²+b² (hyperbola) before computing. This shows the examiner you know the distinction and earns the methodology mark even if arithmetic errors follow.

Model answer

Given: Hyperbola, center at origin O(0,0), transverse axis along x-axis
Vertices: (±4, 0) → a = 4
Foci: (±5, 0) → c = 5
Required: Equation, eccentricity, asymptotes

Step 1 — Identify a and c:
a = 4, c = 5

Step 2 — Compute b using hyperbola relation c² = a² + b²:
5² = 4² + b²
25 = 16 + b²
b² = 9 → b = 3

Step 3 — Write standard equation of hyperbola:
x²/a² − y²/b² = 1
x²/16 − y²/9 = 1

Step 4 — Eccentricity:
e = c/a = 5/4 = 1.25
(Confirmed: e > 1 for a hyperbola ✓)

Step 5 — Equations of asymptotes (y = ±(b/a)x):
y = ±(3/4)x
i.e., 3x − 4y = 0 and 3x + 4y = 0

∴ Equation: x²/16 − y²/9 = 1
Eccentricity: e = 1.25
Asymptotes: y = ±(3/4)x

Answer structure

• Step 1: Identify a = 4 from vertices, c = 5 from foci. [1 mark]
• Step 2: Use c² = a² + b² to find b² = 9. [1 mark]
• Step 3: Write standard equation x²/16 − y²/9 = 1. [1 mark]
• Step 4: Compute eccentricity e = c/a = 5/4. [1 mark]
• Step 5: Write asymptote equations y = ±(b/a)x = ±(3/4)x. [1 mark]

Scoring breakdown

Common mark deductions

• Using c² = a² − b² (ellipse formula) for a hyperbola — this is the most common and most penalized error.
• Writing the asymptote as y = ±(a/b)x instead of y = ±(b/a)x.
• Placing the minus sign under x² instead of y² when the transverse axis is along x.

Key phrases to include

• c² = a² + b²
• x²/a² − y²/b² = 1
• x²/16 − y²/9 = 1
• e = c/a = 5/4
• asymptotes y = ±(b/a)x

Examiner tip

Show the 'add to both sides' step explicitly: (x²−6x+9) + (y²+4y+4) = 3+9+4. This makes the balance of the equation visible to the examiner and earns the method marks even if a subsequent arithmetic slip occurs.

Model answer

Given: x² + y² − 6x + 4y − 3 = 0
Required: Standard form, center (h, k), and radius r

Step 1 — Rewrite, grouping x-terms and y-terms:
(x² − 6x) + (y² + 4y) = 3

Step 2 — Complete the square for x-terms:
x² − 6x + 9 = (x−3)² [add 9 to both sides: half of −6 is −3, (−3)² = 9]

Step 3 — Complete the square for y-terms:
y² + 4y + 4 = (y+2)² [add 4 to both sides: half of 4 is 2, 2² = 4]

Step 4 — Rewrite with completed squares:
(x−3)² + (y+2)² = 3 + 9 + 4
(x−3)² + (y+2)² = 16

Step 5 — Identify center and radius:
Center: (h, k) = (3, −2)
r² = 16 → r = 4

∴ Standard form: (x−3)² + (y+2)² = 16
Center: (3, −2), Radius: r = 4

Answer structure

• Step 1: Group x-terms and y-terms, move constant to RHS. [1 mark]
• Step 2: Complete the square for x: add (6/2)² = 9. [1 mark]
• Step 3: Complete the square for y: add (4/2)² = 4. [1 mark]
• Step 4: Write standard form (x−3)² + (y+2)² = 16. [1 mark]
• Step 5: State center (3, −2) and r = 4. [1 mark]

Scoring breakdown

Common mark deductions

• Adding the completing-the-square values to only the left-hand side — must add to both sides.
• Sign error in center: writing center as (−3, +2) instead of (+3, −2).
• Taking r = 16 instead of r = √16 = 4.
• Forgetting to add both 9 and 4 to the RHS: writing 3 + 9 only, ignoring the 4.

Key phrases to include

• completing the square
• (x−3)² + (y+2)² = 16
• center (3, −2)
• radius r = 4
• add the same quantity to both sides

Examiner tip

Memorize the four standard parabola orientations: opens right y²=4ax, opens left y²=−4ax, opens up x²=4ay, opens down x²=−4ay. The coefficient 4a is always present — a alone is only the focal distance.

Model answer

Given: Vertex at V(0,0), opens right, Focus F(3, 0)
Required: Standard equation of parabola and directrix

Step 1 — Identify a (focal distance = distance from vertex to focus):
a = 3

Step 2 — Standard form for parabola opening right, vertex at origin:
y² = 4ax
y² = 4(3)x
y² = 12x

Step 3 — Directrix (a units to the left of vertex, opposite the focus):
x = −a = −3
Directrix: x = −3

∴ Equation: y² = 12x; Directrix: x = −3

Answer structure

• Step 1: Identify a = 3 from given focus location. [1 mark]
• Step 2: Write y² = 4ax = 12x. [1 mark]
• Step 3: State directrix x = −3. [1 mark]

Scoring breakdown

Common mark deductions

• Confusing 4a with a — writing y² = 3x instead of y² = 12x.
• Writing directrix x = +3 instead of x = −3 (directrix is on the opposite side of the focus).
• Using x² = 4ay (wrong orientation — that opens upward).

Key phrases to include

• y² = 4ax
• a = 3 (focal distance)
• y² = 12x
• directrix x = −a = −3

Examiner tip

For 'show that' or 'prove' questions, you must end with an explicit conclusion sentence. Writing the product alone without stating 'therefore perpendicular' may cost the conclusion mark.

Model answer

Given: L₁: 2x − y + 3 = 0 and L₂: x + 2y − 5 = 0
Required: Prove L₁ ⊥ L₂

Step 1 — Find slopes from general form (m = −A/B):
L₁: A = 2, B = −1 → m₁ = −2/(−1) = 2
L₂: A = 1, B = 2 → m₂ = −1/2

Step 2 — Apply perpendicularity condition:
m₁ × m₂ = 2 × (−1/2) = −1 ✓

∴ Since m₁ × m₂ = −1, lines L₁ and L₂ are perpendicular.

Answer structure

• Step 1: Compute m₁ = 2 and m₂ = −1/2 from line equations. [1 mark]
• Step 2: Show m₁ × m₂ = −1, conclude perpendicularity. [1 mark]

Scoring breakdown

Common mark deductions

• Using m = A/B instead of m = −A/B — sign error.
• Showing the product but not writing the conclusion statement.
• Not simplifying m = −1/(−1) correctly — writing m = −1 instead of +1.

Key phrases to include

• m = −A/B
• m₁ × m₂ = −1
• perpendicular
• m₁ = 2, m₂ = −1/2

Examiner tip

When the coefficients of x² and y² differ (here 9 and 4), factor them out BEFORE completing the square. The value added to RHS is coefficient × (half the linear term)², not just (half the linear term)². This is the most commonly lost mark in this problem type.

Model answer

Given: 9x² + 4y² − 36x + 8y + 4 = 0
Required: Type of conic and center

Step 1 — Factor out coefficients and group:
9(x² − 4x) + 4(y² + 2y) = −4

Step 2 — Complete the square for x:
x² − 4x + 4 = (x−2)² → add 4 inside → 9(4) = 36 added to RHS
9(x² − 4x + 4) = 9(x−2)²

Step 3 — Complete the square for y:
y² + 2y + 1 = (y+1)² → add 1 inside → 4(1) = 4 added to RHS
4(y² + 2y + 1) = 4(y+1)²

Step 4 — Rewrite with RHS balance:
9(x−2)² + 4(y+1)² = −4 + 36 + 4
9(x−2)² + 4(y+1)² = 36

Step 5 — Divide through by 36:
(x−2)²/4 + (y+1)²/9 = 1

Step 6 — Classify and identify center:
The equation is of the form (x−h)²/b² + (y−k)²/a² = 1 with a² = 9 > b² = 4
→ This is an ELLIPSE with center (h, k) = (2, −1)
(Major axis along y-axis since a² = 9 is under y)

∴ Conic: Ellipse; Center: (2, −1)
Semi-major axis a = 3, semi-minor axis b = 2

Answer structure

• Step 1: Group terms and factor coefficients. [1 mark]
• Step 2–3: Complete the square for both x and y groups. [1 mark]
• Step 4: Correctly balance RHS (−4 + 36 + 4 = 36). [1 mark]
• Step 5: Divide by 36 to get standard form (x−2)²/4 + (y+1)²/9 = 1. [1 mark]
• Step 6: Identify as ellipse with center (2, −1). [1 mark]

Scoring breakdown

Common mark deductions

• Adding 4 to RHS instead of 9×4 = 36 for x-group — not accounting for the factor of 9.
• Similarly adding 1 instead of 4×1 = 4 for y-group.
• Calling the conic a circle because both x² and y² are present — circles require equal coefficients.
• Sign error in center: writing center as (−2, +1).

Key phrases to include

• completing the square
• factor out coefficient before completing the square
• (x−2)²/4 + (y+1)²/9 = 1
• ellipse
• center (2, −1)

Do

• Write the complete formula with all terms (e.g., include absolute value bars if part of formula).
• Use standard notation as taught (d, m, e, a, b, c).
• If a range is asked (like eccentricity), state both bounds.
• Use the ∴ symbol to signal the final answer.

Don't

• Do not write incomplete formulas — half a formula earns zero.
• Do not confuse similar formulas (e.g., ellipse vs hyperbola c² relation).
• Do not waste time deriving — at 1 mark, recall is all that is needed.
• Do not omit units if the answer is a distance.

Do

• Always write 'Given:' and 'Required:' before solving.
• State the formula in its general form before substituting numbers.
• Show the substitution step separately from the simplified result.
• Box or underline the final answer.

Don't

• Do not go directly to the answer without showing the formula — this risks losing the formula mark.
• Do not leave fractions unsimplified as a final answer.
• Do not skip the substitution step even if it seems obvious.
• Do not write the answer without units when distance is asked.

Do

• Label each step: 'Step 1:', 'Step 2:', 'Step 3:' — this makes it easy for the examiner to award marks.
• Write intermediate results clearly — do not chain-compute in one line.
• For two-part questions (e.g., midpoint and distance), separate clearly as 'Part 1' and 'Part 2'.
• Verify your answer if time permits — write 'Check:' and substitute back.

Don't

• Do not combine all three steps into one long expression — individual steps must be visible.
• Do not skip showing completing-the-square additions to both sides.
• Do not write the equation in non-standard form when standard form is required.
• Do not assume the examiner will infer a step you have not written.

Do

• Write a brief plan: identify what sub-results are needed before starting computation.
• Use numbered steps (Step 1 through Step 5) and label what each step finds.
• Show all completing-the-square work explicitly, including what you add to both sides.
• State classification (circle, ellipse, etc.) as an explicit sentence, not just an equation.
• Write a final summary line: '∴ Center: ..., Radius: ...' or equivalent.
• Include a verification step if time permits — examiners reward this.

Don't

• Do not rush to the final answer — 5-mark problems are graded on process, not just result.
• Do not use c² = a² − b² for a hyperbola or c² = a² + b² for an ellipse.
• Do not omit the asymptote equations when a hyperbola problem asks for them.
• Do not leave the equation in non-standard form as the final answer.
• Do not confuse (y+k)² with (y−k)² when completing the square.