Solved Exercises on Conic Sections Applications in Grade 10

Master conic sections applications: parabolas, ellipses, hyperbolas in real-world contexts through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Parabolic Satellite Dish
Exercise 1
A satellite dish has a diameter of 10 feet and a depth of 2 feet. If the receiver is placed at the focus, find the distance from the vertex to the focus.
Definition:

Parabola: Set of all points equidistant from a fixed point (focus) and a fixed line (directrix)

Standard form: \(y^2 = 4px\) (opens right) or \(x^2 = 4py\) (opens up)

Parabola application method:
  1. Identify the orientation of the parabola
  2. Set up the coordinate system with vertex at origin
  3. Use given dimensions to find parameters
  4. Calculate focal length using the standard form
Given
Diameter = 10 ft, Depth = 2 ft
Vertex
(0, 0)
Focus
(0, p)
Step 1: Set up the coordinate system

Place vertex at origin (0,0). Since dish opens upward, use form \(x^2 = 4py\)

Step 2: Determine parabola equation

At edge of dish: x = 5 (half diameter), y = 2 (depth)

Substitute: \(5^2 = 4p(2)\), so \(25 = 8p\)

Step 3: Solve for focal parameter

\(p = \frac{25}{8} = 3.125\) feet

Step 4: Interpret the result

The focus is 3.125 feet above the vertex

Focus is located 3.125 feet from the vertex
Final answer:

The receiver should be placed 3.125 feet from the vertex of the dish

Applied rules:

• Parabolic property: All rays parallel to axis reflect to focus

• Standard form: \(x^2 = 4py\) for vertical parabola

• Focal length: Distance from vertex to focus is p

2 Elliptical Bridge Arch
Exercise 2
An elliptical bridge arch has a span of 100 meters and a maximum height of 30 meters. Find the width of the arch at a height of 20 meters.
Definition:

Ellipse: Set of all points where the sum of distances to two fixed points (foci) is constant

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

Given
Span = 100m, Height = 30m
Semi-major axis
a = 50m
b = 30m
Step 1: Set up the coordinate system

Center at origin. Semi-major axis a = 50m (half span), semi-minor axis b = 30m (height)

Step 2: Write ellipse equation

\(\frac{x^2}{50^2} + \frac{y^2}{30^2} = 1\), so \(\frac{x^2}{2500} + \frac{y^2}{900} = 1\)

Step 3: Find width at height y = 20

\(\frac{x^2}{2500} + \frac{20^2}{900} = 1\)

\(\frac{x^2}{2500} + \frac{400}{900} = 1\)

\(\frac{x^2}{2500} = 1 - \frac{4}{9} = \frac{5}{9}\)

Step 4: Solve for x and find width

\(x^2 = 2500 \times \frac{5}{9} = \frac{12500}{9}\)

\(x = \pm\sqrt{\frac{12500}{9}} = \pm\frac{50\sqrt{5}}{3}\)

Width = \(2|x| = \frac{100\sqrt{5}}{3} \approx 74.5\) meters

Width at height 20m ≈ 74.5 meters
Final answer:

The width of the arch at a height of 20 meters is approximately 74.5 meters

Applied rules:

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

• Symmetry: Ellipse symmetric about both axes

• Parameter identification: a is semi-major, b is semi-minor axis

3 Planetary Orbit
Exercise 3
A planet orbits the sun in an elliptical path with the sun at one focus. If the major axis is 200 million km and eccentricity is 0.1, find the distance from the sun to the farthest point of the orbit.
Definition:

Eccentricity: \(e = \frac{c}{a}\), where c is distance from center to focus, a is semi-major axis

Aphelion: Farthest point from sun (focus), distance = a + c

Given
Major axis = 200×10⁶ km, e = 0.1
Semi-major axis
a = 100×10⁶ km
Distance to focus
c = ae
Step 1: Identify known values

Major axis = 2a = 200×10⁶ km, so a = 100×10⁶ km

Eccentricity e = 0.1

Step 2: Calculate distance from center to focus

Using \(e = \frac{c}{a}\), we get \(c = ae = 100×10^6 × 0.1 = 10×10^6\) km

Step 3: Find aphelion distance

Farthest point (aphelion) = a + c = 100×10⁶ + 10×10⁶ = 110×10⁶ km

Step 4: Interpret the result

The planet is 110 million km from the sun at its farthest point

Aphelion distance = 110 million km
Final answer:

The planet is 110 million kilometers from the sun at its farthest point in the orbit

Applied rules:

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

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

• Aphelion distance: a + c (farthest point)

Conic Sections Applications Guide
\(y^2 = 4px \text{ (parabola)}, \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ (ellipse)}\)
Conic Section Equations
Parabola
\(x^2 = 4py\)
Applications: Satellites, reflectors, projectiles
Ellipse
\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
Applications: Orbits, bridges, acoustics
Hyperbola
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Applications: Navigation, telescopes, cooling towers
Key definitions:

Parabola: Set of points equidistant from focus and directrix

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

Hyperbola: Set of points where difference of distances to foci is constant

Application methodology:
  1. Identify the conic: Recognize real-world scenario
  2. Set coordinates: Place in convenient coordinate system
  3. Use properties: Apply geometric properties of conic
  4. Solve algebraically: Find required measurements
Tip 1: For parabolic reflectors, focus is where energy converges.
Tip 2: Elliptical orbits have the sun at one focus.
Tip 3: Hyperbolic navigation uses time differences from two stations.
Tip 4: Always check that dimensions make physical sense.
Common errors: Confusing axes in ellipse, misplacing focus in parabola, incorrect eccentricity interpretation.
Exam preparation: Memorize standard forms, practice setting up coordinate systems, work with real-world units.
Essential formulas:

• Parabola focus distance: \(x^2 = 4py\), focus at \((0,p)\)

• Ellipse eccentricity: \(e = \frac{c}{a}\), \(c^2 = a^2 - b^2\)

• Ellipse distances: Sum to foci = 2a, Aphelion = a+c, Perihelion = a-c

• Hyperbola: Difference to foci = 2a, \(c^2 = a^2 + b^2\)

Solution: Exercises 4 to 5
4 Cooling Tower
Exercise 4
A cooling tower has a hyperbolic cross-section. The base diameter is 120 meters, the narrowest part is 60 meters wide and 80 meters high, and the top is 100 meters high. Find the equation of the hyperbola.
Definition:

Hyperbola: Set of points where absolute difference of distances to two fixed points (foci) is constant

Standard form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (horizontal) or \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) (vertical)

Given
Base: 120m, Narrowest: 60m at 80m high, Top: 100m
Coordinate system
Center at narrowest point
Equation form
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Step 1: Set up coordinate system

Place center at narrowest part (60m wide), so y-axis runs vertically

At center: width = 60m, so semi-width a = 30m

Step 2: Identify key points

Narrowest point: (±30, 0), Base is 80m below: (±60, -80), Top is 20m above: (±x, 20)

Step 3: Use base point to find b

Point (60, -80) is on hyperbola: \(\frac{60^2}{30^2} - \frac{(-80)^2}{b^2} = 1\)

\(\frac{3600}{900} - \frac{6400}{b^2} = 1\), so \(4 - \frac{6400}{b^2} = 1\)

\(\frac{6400}{b^2} = 3\), thus \(b^2 = \frac{6400}{3}\)

Step 4: Write final equation

\(\frac{x^2}{900} - \frac{y^2}{\frac{6400}{3}} = 1\), or \(\frac{x^2}{900} - \frac{3y^2}{6400} = 1\)

\(\frac{x^2}{900} - \frac{3y^2}{6400} = 1\)
Final answer:

The equation of the hyperbolic cross-section is \(\frac{x^2}{900} - \frac{3y^2}{6400} = 1\)

Applied rules:

• Hyperbola standard form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)

• Coordinate placement: Center at narrowest point

• Point substitution: Points on curve satisfy equation

5 LORAN Navigation
Exercise 5
Two radio stations are 200 km apart. A ship receives signals 0.001 seconds later from station B than from station A. If radio waves travel at 300,000 km/s, find the equation of the hyperbolic path the ship could follow.
Definition:

LORAN Navigation: Uses hyperbolic positioning based on time difference of signals

Hyperbola property: Points with constant difference to two foci lie on hyperbola

Given
Stations 200km apart, time diff = 0.001s, speed = 300,000 km/s
Distance difference
d = speed × time
Hyperbola form
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Step 1: Calculate distance difference

Distance = Speed × Time = 300,000 × 0.001 = 300 km

This means |d₁ - d₂| = 300 km, where d₁, d₂ are distances to stations

Step 2: Set up coordinate system

Place stations on x-axis at (-100, 0) and (100, 0), so 2c = 200, thus c = 100

For hyperbola, |d₁ - d₂| = 2a, so 2a = 300, thus a = 150

Step 3: Find b using hyperbola relationship

For hyperbola: \(c^2 = a^2 + b^2\)

\(100^2 = 150^2 + b^2\)

\(10,000 = 22,500 + b^2\)

\(b^2 = 12,500\)

Step 4: Write hyperbola equation

\(\frac{x^2}{150^2} - \frac{y^2}{12,500} = 1\), or \(\frac{x^2}{22,500} - \frac{y^2}{12,500} = 1\)

\(\frac{x^2}{22,500} - \frac{y^2}{12,500} = 1\)
Final answer:

The ship follows the hyperbolic path \(\frac{x^2}{22,500} - \frac{y^2}{12,500} = 1\)

Applied rules:

• Distance formula: Distance = Speed × Time

• Hyperbola definition: Constant difference to foci

• Hyperbola relationship: \(c^2 = a^2 + b^2\)

Conic Sections Properties and Applications Guide
\(e = \frac{c}{a}, c^2 = a^2 - b^2 \text{ (ellipse)}, c^2 = a^2 + b^2 \text{ (hyperbola)}\)
Eccentricity and Relationships
Key definitions:

Conic sections: Curves obtained by intersecting a cone with a plane

Eccentricity (e): Measure of deviation from circular shape

Focus/Foci: Special points defining the conic section

Real-world application methodology:
  1. Identify scenario: Recognize if parabola, ellipse, or hyperbola applies
  2. Set coordinate system: Use symmetry and known points
  3. Apply geometric properties: Use focus properties, axis relationships
  4. Solve algebraically: Substitute known values and solve
Tip 1: Parabolas reflect energy to focus; ellipses reflect between foci; hyperbolas separate energy.
Tip 2: Eccentricity determines conic type: e=0 (circle), 01 (hyperbola).
Tip 3: In engineering applications, conic shapes optimize structural efficiency and energy focus.
Tip 4: Always verify that calculated dimensions match physical constraints.
Common applications: Satellite dishes (parabola), planetary orbits (ellipse), navigation systems (hyperbola), architectural structures.
Key relationships: For ellipse: \(c^2 = a^2 - b^2\), for hyperbola: \(c^2 = a^2 + b^2\), where c is distance to focus.
Essential formulas:

• Parabola: \(y^2 = 4px\) (focus at \((p,0)\)), \(x^2 = 4py\) (focus at \((0,p)\))

• Ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), eccentricity \(e = \frac{c}{a}\), \(c^2 = a^2 - b^2\)

• Hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), eccentricity \(e = \frac{c}{a}\), \(c^2 = a^2 + b^2\)

• Distance properties: Parabola (point to focus = point to directrix), Ellipse (sum to foci = 2a), Hyperbola (diff to foci = 2a)

Visualization: Conic Section Applications
Conic Section Characteristics
Compare the characteristics of parabolas, ellipses, and hyperbolas:
Parabola: \(y^2 = 4px\), Eccentricity = 1
Ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), Eccentricity < 1
Hyperbola: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), Eccentricity > 1

Analysis: The chart shows how eccentricity varies across conic sections and their practical applications.

  • Parabola (e=1): Used in satellite dishes, flashlights, projectile motion
  • Ellipse (e<1): Used in planetary orbits, whispering galleries, medical lithotripsy
  • Hyperbola (e>1): Used in navigation systems, telescope mirrors, cooling towers

Questions & Answers

Question: I don't understand why parabolic reflectors focus all parallel rays to a single point. Can you explain the mathematical reasoning behind this?

Answer: The focusing property of parabolas is based on the optical law that the angle of incidence equals the angle of reflection.

  • By definition, every point on a parabola is equidistant from the focus and the directrix
  • For a ray parallel to the axis of symmetry, when it hits the parabola, it reflects toward the focus
  • This happens because the tangent line at any point creates equal angles with the line to the focus and the line to the directrix

Mathematically, for parabola \(y^2 = 4px\), any ray parallel to the x-axis hitting the parabola at point \((x,y)\) will reflect along the line connecting that point to the focus \((p,0)\).

This is why satellite dishes, car headlights, and telescopes use parabolic shapes - they collect or project energy efficiently to/from a single point.

Question: When calculating orbital distances, how do I determine whether to use perihelion or aphelion in a problem?

Answer: The choice depends on the specific context of the problem:

  • Perihelion: Closest approach to the Sun (distance = a - c)
  • Aphelion: Farthest distance from the Sun (distance = a + c)
  • Problem keywords: Look for phrases like "closest," "minimum distance," "nearest" for perihelion
  • Problem keywords: Look for "farthest," "maximum distance," "greatest" for aphelion

Example: "Find the closest distance from Earth to the Sun" → Use perihelion = a - c

"Find how far Mars travels from the Sun at its most distant point" → Use aphelion = a + c

Remember: a is semi-major axis, c is distance from center to focus.

Question: Why do cooling towers have hyperbolic shapes? What advantages does this provide?

Answer: Hyperbolic shapes in cooling towers offer several engineering advantages:

  • Structural stability: The hyperboloid shape provides excellent strength-to-weight ratio
  • Material efficiency: Less material needed compared to cylindrical shapes
  • Aerodynamic flow: The narrowing middle section accelerates air flow, improving cooling efficiency
  • Thermal performance: Shape promotes natural draft for heat dissipation

The hyperbolic design also allows for minimal stress concentration and can withstand high wind loads. The mathematical precision of the hyperbola ensures optimal structural performance while maintaining the necessary cooling surface area.

This is a perfect example of how conic sections solve real-world engineering challenges!

Question: How does the LORAN navigation system use hyperbolas to determine position?

Answer: LORAN (Long Range Navigation) uses the mathematical property that points with constant difference in distance to two fixed points lie on a hyperbola.

Here's how it works:

  • Two radio stations transmit synchronized signals
  • The receiver measures the time difference between receiving the two signals
  • This time difference corresponds to a distance difference: Δd = speed × Δt
  • All positions with this distance difference form a hyperbolic curve
  • Using multiple pairs of stations creates intersecting hyperbolas
  • The intersection point gives the precise location

This system was widely used before GPS and demonstrates how hyperbolas solve real navigation problems!

Question: How do I decide which conic section to use for a particular real-world problem?

Answer: Here's a systematic approach to identify the correct conic section:

  1. Parabola: Look for focusing properties, reflection of parallel rays, or projectile motion
  2. Ellipse: Look for closed orbits, sum of distances being constant, or bounded paths
  3. Hyperbola: Look for open paths, difference of distances being constant, or separation properties

Specific indicators:

  • Parabola: Satellite dishes, spotlights, ballistics
  • Ellipse: Planetary orbits, whispering galleries, medical equipment
  • Hyperbola: Navigation systems, cooling towers, lens design

Always consider the geometric property involved: equal distances (parabola), sum of distances (ellipse), or difference of distances (hyperbola).