Solved Exercises on Standard Forms of Ellipse in Grade 1

Master standard forms of ellipse: horizontal and vertical ellipses, center, vertices, foci, and eccentricity through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Horizontal ellipse
Exercise 1
Find the center, vertices, and foci of the ellipse:
\(\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1\)
Definition:

Horizontal ellipse: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where \(a > b\)

Ellipse analysis method:
  1. Identify the standard form and parameters (h, k, a, b)
  2. Determine if it's horizontal or vertical
  3. Calculate c using \(c^2 = a^2 - b^2\)
  4. Find center, vertices, and foci
Given equation
\(\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1\)
Parameters
h=2, k=-3, a=5, b=3
Step 1: Identify parameters

Comparing with standard form: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)

Center: (h, k) = (2, -3)

a² = 25 → a = 5, b² = 9 → b = 3

Step 2: Determine orientation

Since a > b, this is a horizontal ellipse

Major axis length = 2a = 10, Minor axis length = 2b = 6

Step 3: Calculate focal distance

Using formula: \(c^2 = a^2 - b^2\)

c² = 25 - 9 = 16 → c = 4

Step 4: Find vertices and foci

Vertices: (h±a, k) = (2±5, -3) = (7, -3) and (-3, -3)

Foci: (h±c, k) = (2±4, -3) = (6, -3) and (-2, -3)

Center: (2, -3), Vertices: (7, -3), (-3, -3), Foci: (6, -3), (-2, -3)
Final answer:

Center: (2, -3), Vertices: (7, -3) and (-3, -3), Foci: (6, -3) and (-2, -3)

Applied rules:

• Standard form identification: Match coefficients to find h, k, a, b

• Orientation rule: If a > b, horizontal ellipse; if b > a, vertical ellipse

• Focal relationship: \(c^2 = a^2 - b^2\) for ellipses

2 Vertical ellipse
Exercise 2
Find the center, vertices, and foci of the ellipse:
\(\frac{(x+1)^2}{4} + \frac{(y-2)^2}{16} = 1\)
Definition:

Vertical ellipse: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) where \(a > b\)

Given equation
\(\frac{(x+1)^2}{4} + \frac{(y-2)^2}{16} = 1\)
Parameters
h=-1, k=2, a=4, b=2
Focal distance
c=2√3 ≈ 3.46
Step 1: Identify parameters

Comparing with standard form: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)

Center: (h, k) = (-1, 2)

a² = 16 → a = 4, b² = 4 → b = 2

Step 2: Determine orientation

Since a > b and the larger denominator is under y-term, this is a vertical ellipse

Major axis length = 2a = 8, Minor axis length = 2b = 4

Step 3: Calculate focal distance

Using formula: \(c^2 = a^2 - b^2\)

c² = 16 - 4 = 12 → c = √12 = 2√3 ≈ 3.46

Step 4: Find vertices and foci

For vertical ellipse: Vertices: (h, k±a) = (-1, 2±4) = (-1, 6) and (-1, -2)

Foci: (h, k±c) = (-1, 2±2√3) = (-1, 2+2√3) and (-1, 2-2√3)

Center: (-1, 2), Vertices: (-1, 6), (-1, -2), Foci: (-1, 2+2√3), (-1, 2-2√3)
Final answer:

Center: (-1, 2), Vertices: (-1, 6) and (-1, -2), Foci: (-1, 2+2√3) and (-1, 2-2√3)

Applied rules:

• Vertical orientation: Larger denominator under y-term indicates vertical ellipse

• Vertex calculation: For vertical ellipse, vertices are at (h, k±a)

• Focal positioning: Foci are aligned along the major axis

3 Eccentricity calculation
Exercise 3
Find the eccentricity of the ellipse from Exercise 1:
\(\frac{(x-2)^2}{25} + \frac{(y+3)^2}{9} = 1\)
Definition:

Eccentricity: \(e = \frac{c}{a}\) where 0 < e < 1 for ellipses

From Ex1
a=5, c=4
Formula
e = c/a
Result
e = 4/5 = 0.8
Step 1: Recall values from Exercise 1

From previous exercise: a = 5, c = 4

Step 2: Apply eccentricity formula

Eccentricity formula: \(e = \frac{c}{a}\)

\(e = \frac{4}{5} = 0.8\)

Step 3: Interpret the result

Since e = 0.8 (close to 1), this ellipse is relatively elongated

If e were closer to 0, it would be more circular

Step 4: Understand eccentricity meaning

Eccentricity measures how much an ellipse deviates from being circular

e = 0: circle, 0 < e < 1: ellipse, e approaching 1: very elongated ellipse

e = 4/5 = 0.8 (relatively elongated ellipse)
Final answer:

The eccentricity is e = 0.8, indicating a relatively elongated ellipse

Applied rules:

• Eccentricity range: For ellipses, 0 < e < 1

• Shape interpretation: Higher e means more elongated shape

• Focal relationship: \(c^2 = a^2 - b^2\) connects all ellipse parameters

Rules and methods, laws,...
\(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
Horizontal Ellipse
Horizontal
\(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
Major axis: horizontal
Vertical
\(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)
Major axis: vertical
Eccentricity
\(e = \frac{c}{a}, c^2 = a^2 - b^2\)
Shape measure
Key definitions:

Ellipse: Set of all points where sum of distances to two foci is constant

Center: Midpoint of line segment joining foci (h, k)

Vertices: Points where ellipse intersects major axis

Foci: Two fixed points inside ellipse determining its shape

Major axis: Longest diameter of ellipse

Minor axis: Shortest diameter of ellipse

Complete methodology:
  1. Identify standard form: Recognize whether horizontal or vertical ellipse
  2. Extract parameters: Find h, k, a, b from equation
  3. Determine orientation: Compare a and b to find major axis
  4. Calculate focal distance: Use \(c^2 = a^2 - b^2\)
  5. Find geometric features: Center, vertices, foci positions
  6. Calculate eccentricity: Use \(e = c/a\)
Tip 1: Always identify whether it's horizontal or vertical by checking which denominator is larger.
Tip 2: Remember a is always associated with the major axis (longer direction).
Tip 3: For eccentricity, c is always less than a in an ellipse (0 < e < 1).
Tip 4: The sum of distances from any point on ellipse to both foci equals 2a.
Common errors: Confusing a and b roles, misidentifying major axis, forgetting to take square roots.
Exam preparation: Memorize both standard forms, practice parameter identification, master eccentricity calculations.
Formulas to know by heart:

• Horizontal ellipse: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) (a > b)

• Vertical ellipse: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) (a > b)

• Relationship: \(c^2 = a^2 - b^2\)

• Eccentricity: \(e = \frac{c}{a}\)

• Vertices: (h±a, k) or (h, k±a) depending on orientation

• Foci: (h±c, k) or (h, k±c) depending on orientation

Solution: Exercises 4 to 5
4 Converting to standard form
Exercise 4
Convert to standard form and find center, vertices:
\(4x^2 + 9y^2 - 16x + 18y - 11 = 0\)
Definition:

Completing the square: Method to convert general form to standard form

Original
\(4x^2 + 9y^2 - 16x + 18y - 11 = 0\)
Group
\(4(x^2 - 4x) + 9(y^2 + 2y) = 11\)
Complete
\(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1\)
Step 1: Group x and y terms

Rearrange: \(4x^2 - 16x + 9y^2 + 18y = 11\)

Factor coefficients: \(4(x^2 - 4x) + 9(y^2 + 2y) = 11\)

Step 2: Complete the square for x terms

For \(x^2 - 4x\): Take half of -4 (which is -2), square it: (-2)² = 4

\(x^2 - 4x + 4 - 4 = (x-2)^2 - 4\)

So: \(4[(x-2)^2 - 4] = 4(x-2)^2 - 16\)

Step 3: Complete the square for y terms

For \(y^2 + 2y\): Take half of 2 (which is 1), square it: 1² = 1

\(y^2 + 2y + 1 - 1 = (y+1)^2 - 1\)

So: \(9[(y+1)^2 - 1] = 9(y+1)^2 - 9\)

Step 4: Substitute back and solve

\(4(x-2)^2 - 16 + 9(y+1)^2 - 9 = 11\)

\(4(x-2)^2 + 9(y+1)^2 = 36\)

Divide by 36: \(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1\)

Step 5: Identify parameters

Standard form: \(\frac{(x-2)^2}{3^2} + \frac{(y+1)^2}{2^2} = 1\)

Center: (2, -1), a = 3, b = 2

Since a > b, horizontal ellipse

Step 6: Find vertices

Vertices: (h±a, k) = (2±3, -1) = (5, -1) and (-1, -1)

Standard form: \(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1\), Center: (2, -1), Vertices: (5, -1), (-1, -1)
Final answer:

Standard form: \(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1\), Center: (2, -1), Vertices: (5, -1) and (-1, -1)

Applied rules:

• Completing the square: Add and subtract \((\frac{coefficient}{2})^2\)

• Factoring: Factor out coefficients before completing the square

• Normalization: Divide by constant to get 1 on right side

5 Real-world application
Exercise 5
An elliptical garden has equation \(\frac{x^2}{100} + \frac{y^2}{64} = 1\). Find dimensions and fencing needed.
Definition:

Real-world ellipse: Major/minor axes represent actual dimensions

Equation
\(\frac{x^2}{100} + \frac{y^2}{64} = 1\)
Parameters
a=10, b=8
Dimensions
20×16 ft
Step 1: Identify parameters

Standard form: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

a² = 100 → a = 10, b² = 64 → b = 8

Step 2: Calculate dimensions

Major axis length = 2a = 20 feet (length of garden)

Minor axis length = 2b = 16 feet (width of garden)

Step 3: Calculate perimeter approximation

Perimeter ≈ π[a + b] = π[10 + 8] = 18π ≈ 56.55 feet

(Exact ellipse perimeter requires calculus)

Step 4: Additional features

Area = πab = π(10)(8) = 80π ≈ 251.33 sq ft

Focal distance: c² = 100 - 64 = 36, so c = 6

Step 5: Fencing requirement

About 56.55 feet of fencing needed for perimeter

Foci are 6 feet from center along major axis

Dimensions: 20ft × 16ft, Area: 80π sq ft, Fencing: ~56.55 ft
Final answer:

Garden is 20 feet long and 16 feet wide, requiring approximately 56.55 feet of fencing

Applied rules:

• Dimension interpretation: Axes lengths are 2a and 2b

• Area formula: Area = πab for ellipse

• Approximation: Perimeter ≈ π(a+b) for practical purposes

Ellipse Properties and Applications Guide
\(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
Standard Ellipse Equation
Key definitions:

Ellipse: Locus of points where sum of distances to two fixed points (foci) is constant

Focus/Foci: Two fixed points inside ellipse that determine its shape

Major axis: Longest diameter passing through both foci

Minor axis: Shortest diameter perpendicular to major axis

Eccentricity: Measure of deviation from circularity (0 < e < 1)

Complete methodology:
  1. Pattern recognition: Identify standard form and orientation
  2. Parameter extraction: Carefully identify h, k, a, b
  3. Geometric calculation: Use relationships to find vertices, foci
  4. Verification: Check results against original equation
Tip 1: Remember that a is always associated with the major axis, regardless of orientation.
Tip 2: In standard form, denominators are always positive and represent squares of semi-axes.
Tip 3: The eccentricity tells you how "stretched" the ellipse is compared to a circle.
Tip 4: For conversion problems, group terms first, then complete the square for each variable.
Common applications: Planetary orbits, architectural arches, acoustic properties, engineering designs.
Key relationships: c² = a² - b², e = c/a, sum of distances to foci = 2a.
Essential formulas:

• Horizontal ellipse: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) (a > b)

• Vertical ellipse: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) (a > b)

• Focal relationship: \(c^2 = a^2 - b^2\)

• Eccentricity: \(e = \frac{c}{a}\)

• Area: \(A = \pi ab\)

• Approximate perimeter: \(P \approx \pi(a+b)\)

Exercise with Visualization: Ellipse Families
Exercise 6: Ellipse Parameter Effects
Analyze how changing parameters affects ellipse shape:
\(E_1: \frac{x^2}{25} + \frac{y^2}{9} = 1\) (horizontal)
\(E_2: \frac{x^2}{9} + \frac{y^2}{25} = 1\) (vertical)
\(E_3: \frac{(x-2)^2}{16} + \frac{(y+1)^2}{4} = 1\) (translated)

Analysis: The visualization shows how parameters affect ellipse orientation and position.

  • \(E_1\): Horizontal ellipse centered at origin, a=5, b=3
  • \(E_2\): Vertical ellipse centered at origin, a=5, b=3
  • \(E_3\): Horizontal ellipse centered at (2,-1), a=4, b=2

Questions & Answers

Question: I don't understand why in the standard form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), sometimes a is under the x-term and sometimes under the y-term. How do I know which is which?

Answer: Great question! The key is understanding that 'a' is always associated with the MAJOR axis (the longer one), regardless of orientation:

  • When the major axis is horizontal (wider than tall): \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) where a > b
  • When the major axis is vertical (taller than wide): \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\) where a > b

To identify which is which:

  1. Look at the denominators: the larger denominator corresponds to 'a²'
  2. Whatever variable (x or y) has the larger denominator determines the orientation
  3. 'a' is always the semi-major axis length, 'b' is the semi-minor axis length

Example: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\) → 25 > 9, so a² = 25 (a = 5) under x-term → horizontal ellipse

Question: When converting from general form to standard form, I get confused with completing the square. Can you explain the process more clearly?

Answer: Completing the square follows a systematic approach:

  1. Group like terms: Collect all x terms together and all y terms together
  2. Factor coefficients: If x² or y² have coefficients other than 1, factor them out from their respective groups
  3. Complete the square: For x terms in form x² + px, add and subtract (p/2)²
  4. Balance equation: Whatever you add to one side, add to the other (remember factored coefficients!)
  5. Factor perfect squares: Write completed squares as (x - h)² and (y - k)²
  6. Normalize: Divide everything by the constant on the right to get 1

Example with 4x² + 9y² - 16x + 18y - 11 = 0:

  • Group: 4x² - 16x + 9y² + 18y = 11
  • Factor: 4(x² - 4x) + 9(y² + 2y) = 11
  • Complete: 4[(x² - 4x + 4) - 4] + 9[(y² + 2y + 1) - 1] = 11
  • Simplify: 4(x - 2)² - 16 + 9(y + 1)² - 9 = 11
  • Combine: 4(x - 2)² + 9(y + 1)² = 36
  • Normalize: \(\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1\)

Question: What does eccentricity actually mean? Why is it important in describing ellipses?

Answer: Eccentricity (e) is a dimensionless number that measures how much an ellipse deviates from being circular:

  • e = 0: Perfect circle (both foci at center)
  • 0 < e < 1: Ellipse (the closer to 1, the more elongated)
  • e = 1: Parabola (degenerate case)

The formula e = c/a relates the focal distance to the semi-major axis. Since c = √(a² - b²), we have:

  • When a ≈ b (nearly circular): c is small, so e is small
  • When a >> b (very elongated): c approaches a, so e approaches 1

Practical importance:

  • Planetary orbits: Earth's orbit has e ≈ 0.017 (nearly circular)
  • Engineering: Design of reflectors and acoustic chambers
  • Physics: Understanding orbital mechanics and gravitational fields

Question: How can I remember all the different formulas for ellipses? There seem to be many relationships.

Answer: Focus on the core relationships and derive others from them:

  1. Start with definition: Sum of distances to foci is constant = 2a
  2. Remember Pythagorean relationship: c² = a² - b² (like right triangle with hypotenuse a)
  3. Derive eccentricity: e = c/a (ratio of focal distance to semi-major axis)

Memory aids:

  • "a is always biggest": a ≥ b, and a is always associated with the major axis
  • "c is in the middle": c² = a² - b², so c is between a and b
  • "eccentricity is small c over big a": e = c/a

For orientation:

  • Horizontal: Larger denominator under x-term: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
  • Vertical: Larger denominator under y-term: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)

Practice with visual representations to reinforce the geometric meanings!

Question: Are there any real-world applications of ellipses beyond just mathematical theory?

Answer: Ellipses have numerous practical applications across various fields:

  1. Astronomy: Planetary orbits are elliptical (Kepler's laws), with the sun at one focus
  2. Architecture: Elliptical arches distribute weight efficiently, used in bridges and buildings
  3. Acoustics: Whispering galleries use elliptical domes where sound from one focus reaches the other
  4. Medicine: Lithotripsy uses elliptical reflectors to focus shock waves for kidney stone treatment
  5. Engineering: Gears, optical systems, and satellite trajectories utilize elliptical properties

Special properties that make ellipses valuable:

  • Reflection property: Light/sound from one focus reflects to the other focus
  • Constant sum property: Useful in optimization problems
  • Smooth curves: Ideal for aerodynamic and hydrodynamic applications

Understanding ellipse properties helps engineers and scientists optimize designs and predict natural phenomena!