Online Self Test
Learn & Practice Quizzes & Challenges Competitive Exams
 

Conic Sections – CBSE NCERT Study Resources

Previous ChapterNext Chapter
Previous Chapter Next Chapter

Study Materials

11th

11th - Mathematics

Conic Sections

Jump to Question Sets

Overview of the Chapter: Conic Sections

This chapter introduces the concept of conic sections, which are curves obtained by intersecting a right circular cone with a plane. These curves include circles, ellipses, parabolas, and hyperbolas, each with distinct geometric properties and equations. The chapter explores their standard forms, properties, and applications in mathematics and real-world scenarios.

Sections of a Cone

Conic sections are formed when a plane intersects a double-napped right circular cone. Depending on the angle of intersection, different curves are obtained:

  • Circle: Formed when the plane cuts the cone parallel to its base.
  • Parabola: Formed when the plane is parallel to the generator of the cone.
  • Ellipse: Formed when the plane cuts the cone obliquely, but not parallel to the base or generator.
  • Hyperbola: Formed when the plane cuts both nappes of the cone.

Definition: A conic section is the locus of a point moving in a plane such that the ratio of its distance from a fixed point (focus) to its distance from a fixed line (directrix) is constant (eccentricity).

Circle

A circle is the set of all points in a plane equidistant from a fixed point (center). Its standard equation is:

(x - h)2 + (y - k)2 = r2

where (h, k) is the center and r is the radius.

Parabola

A parabola is the set of points equidistant from a fixed point (focus) and a fixed line (directrix). Its standard forms are:

  • y2 = 4ax (opens right)
  • y2 = -4ax (opens left)
  • x2 = 4ay (opens upward)
  • x2 = -4ay (opens downward)

Ellipse

An ellipse is the set of points where the sum of distances from two fixed points (foci) is constant. Its standard equation is:

x2/a2 + y2/b2 = 1

where a > b, and the major axis is 2a.

Hyperbola

A hyperbola is the set of points where the difference of distances from two fixed points (foci) is constant. Its standard equation is:

x2/a2 - y2/b2 = 1

where the transverse axis is 2a.

Applications of Conic Sections

Conic sections have wide applications in physics, engineering, astronomy, and architecture. For example:

  • Parabolas are used in satellite dishes and headlights.
  • Ellipses describe planetary orbits.
  • Hyperbolas are used in navigation systems.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define a parabola as a conic section.
Answer:

A parabola is the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix).

Question 2:
What is the standard equation of a circle centered at (h, k)?
Answer:

(x - h)² + (y - k)² = r²

Question 3:
Identify the conic section represented by 4x² + 9y² = 36.
Answer:

Ellipse (standard form: x²/9 + y²/4 = 1).

Question 4:
What is the eccentricity of a circle?
Answer:
0
Question 5:
Find the focus of the parabola y² = 12x.
Answer:

(3, 0)

Question 6:
Write the condition for a general second-degree equation to represent a hyperbola.
Answer:

Δ ≠ 0 and h² > ab (where Δ = discriminant).

Question 7:
What are the vertices of the ellipse x²/25 + y²/16 = 1?
Answer:

(±5, 0)

Question 8:
Give the equation of the directrix for the parabola x² = -8y.
Answer:

y = 2

Question 9:
What is the length of the latus rectum of the hyperbola x²/9 - y²/4 = 1?
Answer:

8/3 units

Question 10:
Identify the conic section with eccentricity e = 1.5.
Answer:

Hyperbola (since e > 1).

Question 11:
Find the center of the hyperbola (y-3)²/16 - (x+2)²/9 = 1.
Answer:

(-2, 3)

Question 12:
What is the angle between asymptotes of the hyperbola xy = 8?
Answer:

90° (rectangular hyperbola).

Question 13:
What is the standard equation of a circle with center at (h, k) and radius r?
Answer:

The standard equation of a circle is:
(x - h)2 + (y - k)2 = r2
where (h, k) is the center and r is the radius.

Question 14:
Identify the conic section represented by the equation 4x2 + 9y2 = 36.
Answer:

The equation 4x2 + 9y2 = 36 represents an ellipse because it can be rewritten as:
x2/9 + y2/4 = 1,
which is the standard form of an ellipse.

Question 15:
Write the condition for the general equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 to represent a hyperbola.
Answer:

The condition for the equation to represent a hyperbola is:
B2 - 4AC > 0.
This is derived from the discriminant of the conic section.

Question 16:
Find the coordinates of the focus of the parabola y2 = 8x.
Answer:

For the parabola y2 = 8x, the standard form is y2 = 4ax.
Comparing, 4a = 8 ⇒ a = 2.
Thus, the focus is at (a, 0) = (2, 0).

Question 17:
What is the length of the latus rectum of the ellipse x2/16 + y2/9 = 1?
Answer:

For the ellipse x2/16 + y2/9 = 1,
a2 = 16 ⇒ a = 4,
b2 = 9 ⇒ b = 3.
The length of the latus rectum is 2b2/a = 2(9)/4 = 4.5 units.

Question 18:
State the equation of the directrix for the parabola x2 = -12y.
Answer:

For the parabola x2 = -12y, the standard form is x2 = -4ay.
Comparing, 4a = 12 ⇒ a = 3.
The directrix is y = a ⇒ y = 3.

Question 19:
What is the transverse axis of the hyperbola x2/9 - y2/4 = 1?
Answer:

For the hyperbola x2/9 - y2/4 = 1, the transverse axis is along the x-axis because the x2 term is positive.

Question 20:
Find the center of the hyperbola (y - 3)2/16 - (x + 2)2/25 = 1.
Answer:

The center of the hyperbola (y - k)2/a2 - (x - h)2/b2 = 1 is at (h, k).
Here, h = -2 and k = 3.
Thus, the center is at (-2, 3).

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Define a parabola and write its standard equation when the vertex is at the origin and the focus lies on the positive x-axis.
Answer:

A parabola is the set of all points in a plane equidistant from a fixed point (focus) and a fixed line (directrix).
Standard equation when vertex is at origin and focus on positive x-axis:
y² = 4ax, where 'a' is the distance from vertex to focus.

Question 2:
What is the eccentricity of a circle? Justify your answer.
Answer:

The eccentricity (e) of a circle is 0.
Reason: For a circle, the distance from center to any point on circumference (radius) is constant, making the ratio of distances (focus to point)/(directrix to point) zero.

Question 3:
Write the condition for the general second degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 to represent a hyperbola.
Answer:

The equation represents a hyperbola if B² - 4AC > 0.
This is because the discriminant determines the conic section type, and positive discriminant corresponds to hyperbola.

Question 4:
Find the coordinates of foci of the ellipse x²/25 + y²/9 = 1.
Answer:

Given ellipse equation: x²/25 + y²/9 = 1
Here, a² = 25 ⇒ a = 5
b² = 9 ⇒ b = 3
Foci distance c = √(a² - b²) = √(25-9) = 4
Foci coordinates: (±4, 0)

Question 5:
What is the length of the latus rectum of the parabola y² = 12x?
Answer:

Given parabola: y² = 12x
Comparing with standard form y² = 4ax:
4a = 12 ⇒ a = 3
Length of latus rectum = 4a = 12 units

Question 6:
Differentiate between a horizontal and vertical hyperbola based on their standard equations.
Answer:
  • Horizontal hyperbola: Standard equation is x²/a² - y²/b² = 1
    Transverse axis is along x-axis
  • Vertical hyperbola: Standard equation is y²/a² - x²/b² = 1
    Transverse axis is along y-axis
Question 7:
Find the center and radius of the circle given by the equation x² + y² - 6x + 4y - 12 = 0.
Answer:

Rewrite equation: x² - 6x + y² + 4y = 12
Complete squares:
(x² - 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x-3)² + (y+2)² = 25
Center: (3, -2)
Radius: √25 = 5 units

Question 8:
What are the parametric equations of a circle with radius 'r' centered at the origin?
Answer:

Parametric equations of circle:
x = r cosθ
y = r sinθ
where θ is the parameter (angle in radians) measured from positive x-axis.

Question 9:
For the hyperbola 9x² - 16y² = 144, find the equations of its asymptotes.
Answer:

Rewrite equation: x²/16 - y²/9 = 1
Asymptotes for standard hyperbola x²/a² - y²/b² = 1 are y = ±(b/a)x
Here a=4, b=3
Asymptotes: y = ±(3/4)x

Question 10:
If the distance between foci of an ellipse is 10 and minor axis length is 8, find its major axis length.
Answer:

Given: Distance between foci = 2c = 10 ⇒ c = 5
Minor axis length = 2b = 8 ⇒ b = 4
Using c² = a² - b²:
25 = a² - 16 ⇒ a² = 41 ⇒ a = √41
Major axis length = 2a = 2√41

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define a parabola and derive its standard equation in the form y² = 4ax.
Answer:

A parabola is the locus of a point that moves in a plane such that its distance from a fixed point (called the focus) is equal to its distance from a fixed line (called the directrix).


To derive the standard equation y² = 4ax:
1. Let the focus be at (a, 0) and the directrix be the line x = -a.
2. For any point P(x, y) on the parabola, distance to focus = distance to directrix.
3. Using distance formula: √[(x - a)² + y²] = |x + a|
4. Squaring both sides: (x - a)² + y² = (x + a)²
5. Simplifying: y² = 4ax
Question 2:
Differentiate between the equations of an ellipse and a hyperbola in their standard forms.
Answer:

The standard equations and key differences are:

  • Ellipse: (x²/a²) + (y²/b²) = 1 (where a > b)
  • Hyperbola: (x²/a²) - (y²/b²) = 1

Key differences:

  • Ellipse has a sum of distances from two foci constant, hyperbola has a difference.
  • Ellipse is a closed curve, hyperbola has two open branches.
  • For ellipse, b² = a²(1 - e²); for hyperbola, b² = a²(e² - 1).
Question 3:
Find the equation of the circle with center at (2, -3) and radius 5 units.
Answer:

The standard equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r².


Given:
Center (h, k) = (2, -3)
Radius r = 5

Substituting values:
(x - 2)² + (y - (-3))² = 5²
Simplifying:
(x - 2)² + (y + 3)² = 25
Question 4:
Explain the condition when the general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight lines.
Answer:

The general second degree equation represents a pair of straight lines if the determinant of its coefficients is zero.


The condition is:
| a h g |
| h b f | = 0
| g f c |

Expanding this determinant gives:
abc + 2fgh - af² - bg² - ch² = 0

This is called the degenerate condition for a conic section. When satisfied, the equation can be factored into two linear equations representing two straight lines.

Question 5:
A rod of length 12 cm moves with its ends always touching the coordinate axes. Find the equation of the locus of a point P on the rod which is 3 cm from the end in contact with the x-axis.
Answer:

This problem describes an ellipse as the locus.


Let the rod AB touch axes at A (a,0) and B (0,b).
Given AB = 12 cm ⇒ a² + b² = 144 (Pythagoras theorem)

Point P divides AB in ratio AP:PB = 3:9 = 1:3
Coordinates of P:
x = (3×0 + 1×a)/(1+3) = a/4 ⇒ a = 4x
y = (3×b + 1×0)/(1+3) = 3b/4 ⇒ b = 4y/3

Substituting in a² + b² = 144:
(4x)² + (4y/3)² = 144
16x² + (16y²)/9 = 144
Dividing by 16:
x² + y²/9 = 9/16
Final equation:
(x²)/(9/16) + (y²)/(81/16) = 1
Question 6:
Find the coordinates of the foci and the length of the latus rectum for the ellipse 9x² + 16y² = 144.
Answer:

First, rewrite the equation in standard form by dividing by 144:
x²/16 + y²/9 = 1.


Here, a² = 16 (a = 4) and b² = 9 (b = 3).
For an ellipse, foci are at (±c, 0), where c = √(a² - b²).
Thus, c = √(16 - 9) = √7.
Foci: (√7, 0) and (-√7, 0).
Length of latus rectum = 2b²/a = 2(9)/4 = 4.5 units.
Question 7:
Differentiate between a circle and a hyperbola based on their standard equations and eccentricity.
Answer:

Circle:
Standard equation: x² + y² = r² (center at origin).
Eccentricity (e) = 0 (since it is a closed curve with constant distance from the center).


Hyperbola:
Standard equation: x²/a² - y²/b² = 1 (opens horizontally).
Eccentricity (e) > 1 (since it is an open curve with diverging branches).


Key difference: A circle has a single radius, while a hyperbola has two distinct branches and asymptotes.
Question 8:
Derive the condition for the line y = mx + c to be tangent to the parabola y² = 4ax.
Answer:

For the line to be tangent to the parabola, the system of equations must have exactly one solution.


Substitute y = mx + c into y² = 4ax:
(mx + c)² = 4ax.
Expand: m²x² + 2mcx + c² - 4ax = 0.
For a single solution, the discriminant (D) must be zero:
(2mc - 4a)² - 4(m²)(c²) = 0.
Simplify: 4m²c² - 16amc + 16a² - 4m²c² = 0.
This reduces to -16amc + 16a² = 0, giving the condition: c = a/m.
Question 9:
A hyperbola has vertices at (±4, 0) and foci at (±5, 0). Write its standard equation and find the length of its transverse axis.
Answer:

Given vertices at (±4, 0), a = 4.
Given foci at (±5, 0), c = 5.
For hyperbolas, c² = a² + b², so b² = 25 - 16 = 9 (b = 3).


Standard equation: x²/16 - y²/9 = 1.
Length of transverse axis = 2a = 8 units (since it is the distance between vertices).

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Derive the standard equation of a parabola with vertex at origin and focus on the positive x-axis. Explain the geometric significance of the latus rectum.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). For a parabola with vertex at origin and focus at (a,0), the directrix is x = -a.

Evidence Analysis
  • Using distance formula: √[(x-a)² + y²] = |x + a|
  • Squaring both sides: (x-a)² + y² = (x+a)²
  • Simplifying gives y² = 4ax (standard equation)
Critical Evaluation

The latus rectum (length 4a) is the chord through focus parallel to directrix. Our textbook shows it determines the parabola's width.

Question 2:
Compare the properties of ellipse and hyperbola using their standard equations. Include one real-world application for each conic section.
Answer:
Theoretical Framework

We studied standard equations: x²/a² + y²/b² = 1 (ellipse) and x²/a² - y²/b² = 1 (hyperbola). Both are central conics with two foci.

Evidence Analysis
  • Ellipse: Sum of distances to foci is constant (2a)
  • Hyperbola: Difference of distances is constant (2a)
  • Eccentricity e < 1 for ellipse, e > 1 for hyperbola
Critical Evaluation

Ellipses model planetary orbits (Kepler's laws). Hyperbolas appear in sonic booms and telescope designs.

Question 3:
Derive the standard equation of a parabola with vertex at origin and focus on the positive x-axis. Explain the significance of the latus rectum in this context.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). For a parabola with vertex at origin and focus at (a,0), the directrix is x = -a.

Evidence Analysis

Using the distance formula, we derive:
√[(x-a)² + y²] = |x + a|
Squaring both sides: (x-a)² + y² = (x+a)²
Simplifying gives y² = 4ax, the standard equation.

Critical Evaluation

The latus rectum (length 4a) is the chord through focus parallel to directrix. Our textbook shows it determines the parabola's width - larger 'a' gives wider parabola.

Question 4:
Compare the properties of ellipse and hyperbola regarding their standard equations, eccentricity, and real-world applications.
Answer:
Theoretical Framework

We studied that both ellipses (x²/a² + y²/b² = 1) and hyperbolas (x²/a² - y²/b² = 1) are conic sections but with different eccentricity (e) ranges: 0≤e<1 for ellipse, e>1 for hyperbola.

Evidence Analysis
  • Ellipse: Sum of distances from foci constant (2a)
  • Hyperbola: Difference of distances constant (2a)
Critical Evaluation

Real-world applications differ: ellipses model planetary orbits (Kepler's laws), while hyperbolas appear in sonic booms and telescope designs due to their asymptotic nature.

Question 5:
Derive the standard equation of a parabola with vertex at origin and focus on the positive x-axis. Explain the significance of the latus rectum in real-world applications like satellite dishes.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from the focus (a,0) and directrix x=-a. Using the distance formula, we derive the standard equation as y²=4ax.


Evidence Analysis
  • Distance from P(x,y) to focus (a,0): √[(x-a)²+y²]
  • Distance to directrix x=-a: |x+a|
  • Equating and squaring yields y²=4ax

Critical Evaluation

The latus rectum (length=4a) determines signal collection efficiency in parabolic reflectors. Our textbook shows how this property maximizes incoming parallel rays at the focus.

Question 6:
Prove that the sum of focal distances of any point on an ellipse is constant and equal to the major axis length. Relate this to planetary orbits using Kepler's laws.
Answer:
Theoretical Framework

For ellipse x²/a² + y²/b²=1 with foci S(ae,0) and S'(-ae,0), we prove PS + PS'=2a for any point P(x,y) on the curve.


Evidence Analysis
  • Using distance formula: PS=√[(x-ae)²+y²]
  • Substitute y² from ellipse equation
  • Simplification shows PS=a-ex, PS'=a+ex
  • Thus PS+PS'=2a (constant)

Future Implications

This mirrors Kepler's 1st law where planets maintain constant total distance to the Sun's two foci, explaining stable orbits.

Question 7:
Compare the properties of ellipse and hyperbola using their standard equations. How do these conic sections model planetary orbits and cooling tower designs respectively?
Answer:
Theoretical Framework

We learned that ellipses (x²/a² + y²/b² = 1) have two foci and closed shape, while hyperbolas (x²/a² - y²/b² = 1) have two branches and open shape.

Evidence Analysis
  • Ellipses model planetary orbits (Kepler's 1st Law) as they maintain constant sum of distances to two foci.
  • Hyperbolas appear in cooling towers due to their structural strength from the curved shape.
Critical Evaluation

The eccentricity (e) determines the conic's shape: e<1 for ellipse, e>1 for hyperbola. This affects their real-world applications.

Future Implications

These properties are crucial in astronomy and engineering designs, showing how abstract math solves practical problems.

Question 8:
Compare the properties of ellipse and hyperbola using their standard equations. Demonstrate how Kepler's laws utilize elliptical orbits with eccentricity calculations.
Answer:
Theoretical Framework

We know ellipses (x²/a² + y²/b² = 1) have 0 ≤ e < 1, while hyperbolas (x²/a² - y²/b² = 1) have e > 1. Both are conic sections with two foci.


Evidence Analysis
  • For ellipses: b² = a²(1-e²). Planets follow elliptical orbits (Kepler's 1st law) with Sun at one focus.
  • Example: Earth's orbit has e ≈ 0.0167 → nearly circular.
  • Hyperbolas model comet trajectories with e > 1, escaping gravitational pull.

Future Implications

Understanding eccentricity helps space agencies calculate orbital insertion and slingshot maneuvers using planetary gravity.

Question 9:
Derive the standard equation of a parabola with vertex at origin and focus on the positive x-axis. Explain the significance of the latus rectum in real-world applications.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). For a vertex at (0,0) and focus at (a,0), the directrix is x = -a.


Evidence Analysis
  • Using distance formula: √[(x-a)² + y²] = |x + a|
  • Squaring both sides: (x-a)² + y² = (x+a)²
  • Simplifies to y² = 4ax (standard equation)

Critical Evaluation

The latus rectum (length 4a) determines the parabola's width. In satellite dishes, this property ensures optimal signal reflection to the focus.

Question 10:
Compare the properties of ellipse and hyperbola using their standard equations. How do these conic sections model planetary orbits and sonic booms respectively?
Answer:
Theoretical Framework

Our textbook shows standard equations: Ellipse (x²/a² + y²/b² = 1) and Hyperbola (x²/a² - y²/b² = 1). Both have two foci but different eccentricity ranges.


Evidence Analysis
  • Ellipse: 0 < e < 1, closed curve (Kepler's 1st law)
  • Hyperbola: e > 1, open curve with asymptotes
  • Planetary orbits are elliptical (Earth's e ≈ 0.0167)
  • Sonic booms trace hyperbolic shockwaves

Future Implications

These models help spacecraft trajectory calculations and noise pollution studies for supersonic aircraft.

Question 11:
Compare the properties of ellipse and hyperbola regarding their eccentricity, foci, and real-world applications. Support with equations from NCERT Class 11.
Answer:
Theoretical Framework

We studied that both ellipses (x²/a² + y²/b² = 1) and hyperbolas (x²/a² - y²/b² = 1) are conic sections but differ fundamentally in eccentricity (e).

Evidence Analysis
  • Ellipse: 0 ≤ e < 1, sum of distances from foci constant
  • Hyperbola: e > 1, difference of distances constant
  • For ellipse: b² = a²(1-e²), for hyperbola: b² = a²(e²-1)
Future Implications

Our textbook shows elliptical orbits (e.g., planets) maintain stable trajectories due to e < 1, while hyperbolas model comet paths (e > 1) that may exit solar systems. These properties are crucial in astrophysics and space mission planning.

Question 12:
Derive the standard equation of a parabola with vertex at origin and focus on the positive x-axis. Explain the significance of the latus rectum in its graphical representation.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). For a parabola with vertex at origin and focus at (a,0), the directrix is x = -a.

Evidence Analysis
  • Using distance formula: √[(x-a)² + y²] = |x + a|
  • Squaring both sides: (x-a)² + y² = (x+a)²
  • Simplifying yields y² = 4ax (standard equation)
Critical Evaluation

The latus rectum (length 4a) is the chord through focus perpendicular to axis. Our textbook shows it determines the parabola's width - larger 'a' gives wider parabola.

Future Implications

This derivation forms basis for analyzing projectile motion in physics, where trajectories follow parabolic paths.

[Diagram: Standard parabola with labeled focus, vertex, and latus rectum]
Question 13:
Compare the properties of ellipse and hyperbola in terms of eccentricity, standard equations, and real-world applications. Support with one textbook example each.
Answer:
Theoretical Framework

We studied that both ellipses (0 < e < 1) and hyperbolas (e > 1) are conic sections, differing in eccentricity 'e'. Their standard equations differ in sign: x²/a² ± y²/b² = 1.

Evidence Analysis
  • Ellipse: x²/9 + y²/4 = 1 (e = √5/3) represents planetary orbits
  • Hyperbola: x²/4 - y²/9 = 1 (e = √13/2) models sonic booms
  • Both have foci, but ellipse is closed curve while hyperbola has two branches
Critical Evaluation

Our textbook shows ellipses model Kepler's laws, while hyperbolas describe navigation systems (LORAN). The sign change in equations creates fundamentally different shapes.

Future Implications

Understanding these differences is crucial for astronomy (ellipses) and physics (hyperbolic trajectories of comets).

[Diagram: Side-by-side comparison of ellipse and hyperbola with labeled parts]
Question 14:
Derive the standard equation of a parabola with vertex at origin and focus at (a, 0). Explain the geometric significance of the parameter a in real-world applications like satellite dishes.
Answer:
Theoretical Framework

We studied that a parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). For a parabola with vertex at origin and focus at (a, 0), the directrix is x = -a.


Evidence Analysis
  • Using distance formula: √[(x-a)² + y²] = |x + a|
  • Squaring both sides: (x-a)² + y² = (x + a)²
  • Simplifying gives y² = 4ax

Critical Evaluation

The parameter a determines the 'width' of the parabola. In satellite dishes, larger a values create deeper dishes for better signal focus, as all incoming parallel rays reflect to the focus.

Question 15:
Compare the eccentricity of ellipse and hyperbola using their standard equations. How does this difference manifest in planetary orbits (ellipses) vs. comet trajectories (hyperbolas)?
Answer:
Theoretical Framework

We learned that eccentricity (e) measures deviation from circularity. For ellipse (x²/a² + y²/b² = 1), e = √(1-b²/a²) < 1. For hyperbola (x²/a² - y²/b² = 1), e = √(1+b²/a²) > 1.


Evidence Analysis
  • Ellipse: 0 < e < 1 (closed orbit)
  • Hyperbola: e > 1 (open trajectory)
  • Circle: e = 0 (special case)

Critical Evaluation

Planets have elliptical orbits (e < 1) due to gravitational binding, while comets may have hyperbolic trajectories (e > 1) if they have escape velocity, never to return.

Question 16:
Define a parabola and derive its standard equation in the form y² = 4ax. Explain the significance of the focus and directrix in its definition.
Answer:

A parabola is the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).


To derive the standard equation y² = 4ax, follow these steps:


1. Assume the focus is at (a, 0) and the directrix is the line x = -a.


2. Let P(x, y) be any point on the parabola. According to the definition, the distance of P from the focus equals its distance from the directrix.


3. Distance from P to focus: √[(x - a)² + y²].


4. Distance from P to directrix: |x + a|.


5. Equate the distances: √[(x - a)² + y²] = |x + a|.


6. Square both sides: (x - a)² + y² = (x + a)².


7. Expand and simplify: x² - 2ax + a² + y² = x² + 2ax + a².


8. Cancel terms: y² = 4ax.


The focus and directrix are crucial as they geometrically define the parabola. The focus lies on the axis of symmetry, and the directrix is perpendicular to it. The vertex, midway between them, is the point where the parabola changes direction.

Question 17:
Compare and contrast the properties of an ellipse and a hyperbola based on their standard equations. Include their eccentricity, foci, and graphical representation.
Answer:

The standard equation of an ellipse is (x²/a²) + (y²/b²) = 1, while a hyperbola has the standard equation (x²/a²) - (y²/b²) = 1.

  • Eccentricity (e):
    - For an ellipse, 0 < e < 1 (measures how much it deviates from a circle).
    - For a hyperbola, e > 1 (indicates an open curve).
  • Foci:
    - Ellipse has two foci inside the curve, at (±ae, 0).
    - Hyperbola also has two foci, but outside the curve, at (±ae, 0).
  • Graphical Representation:
    - Ellipse is a closed, symmetric curve resembling a stretched circle.
    - Hyperbola consists of two open, symmetric branches that extend infinitely.

Both curves are conic sections but differ in shape and properties due to their eccentricity and focal arrangements. The ellipse is bounded, while the hyperbola is unbounded.

Question 18:
Compare and contrast the properties of an ellipse and a hyperbola based on their standard equations. Include the conditions for their eccentricity and the significance of their foci.
Answer:

Comparison of ellipse and hyperbola:

  • Standard Equations:
    Ellipse: x²/a² + y²/b² = 1 (a > b).
    Hyperbola: x²/a² - y²/b² = 1.
  • Eccentricity (e):
    Ellipse: 0 < e < 1 (measures deviation from a circle).
    Hyperbola: e > 1 (indicates open arms).
  • Foci:
    Ellipse: Two fixed points inside the curve (sum of distances from any point is constant).
    Hyperbola: Two fixed points outside the curve (difference of distances from any point is constant).

Both curves are conic sections but differ in shape and properties. The ellipse is a closed curve, while the hyperbola consists of two open branches.

Question 19:
Compare and contrast the properties of an ellipse and a hyperbola with respect to their standard equations, eccentricity, and graphical representation.
Answer:

Ellipse and Hyperbola are both conic sections but differ significantly in properties:


1. Standard Equations:

  • Ellipse: (x²/a²) + (y²/b²) = 1 (if a > b, major axis is horizontal).
  • Hyperbola: (x²/a²) - (y²/b²) = 1 (opens horizontally).

2. Eccentricity (e):

  • Ellipse: 0 ≤ e < 1 (measures deviation from a circle).
  • Hyperbola: e > 1 (indicates open arms of the curve).

3. Graphical Representation:

  • Ellipse: Closed curve with two foci; sum of distances from any point to foci is constant.
  • Hyperbola: Two disjoint branches with two foci; difference of distances from any point to foci is constant.

Both curves have transverse/conjugate axes, but the ellipse is bounded, while the hyperbola extends infinitely. The hyperbola also has asymptotes (y = ±(b/a)x), which the ellipse lacks.

Question 20:
Define a parabola and derive its standard equation in the form y² = 4ax. Also, explain the significance of the focus and directrix in its definition.
Answer:

A parabola is the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

Derivation of the standard equation y² = 4ax:


1. Let the focus be at point S(a, 0) and the directrix be the line x = -a.
2. Consider a point P(x, y) on the parabola. By definition, the distance of P from S equals its distance from the directrix.
3. Distance of P from S: √[(x - a)² + (y - 0)²] = √[(x - a)² + y²].
4. Distance of P from the directrix: |x + a|.
5. Equate the distances: √[(x - a)² + y²] = |x + a|.
6. Square both sides: (x - a)² + y² = (x + a)².
7. Expand and simplify: x² - 2ax + a² + y² = x² + 2ax + a².
8. Cancel terms: y² = 4ax.

Significance of focus and directrix:

  • The focus is a fixed point that helps define the parabola's shape and is crucial for reflecting properties (e.g., parallel rays converge at the focus).
  • The directrix is a fixed line perpendicular to the axis of symmetry, ensuring the equidistant property holds for all points on the parabola.

This equation represents a parabola opening to the right with vertex at the origin, focus at (a, 0), and directrix x = -a.

Question 21:
Define a parabola as a conic section and derive its standard equation in the form y² = 4ax. Include a labeled diagram and explain the significance of the focus and directrix in its definition.
Answer:

A parabola is the locus of a point that moves in a plane such that its distance from a fixed point (called the focus) is equal to its perpendicular distance from a fixed line (called the directrix).

Derivation of the standard equation (y² = 4ax):


1. Let the focus be at point S(a, 0) and the directrix be the line x = -a.
2. Consider a point P(x, y) on the parabola. By definition, PS = PM, where PM is the perpendicular distance from P to the directrix.
3. Using the distance formula: √[(x - a)² + y²] = |x + a|.
4. Squaring both sides: (x - a)² + y² = (x + a)².
5. Expanding: x² - 2ax + a² + y² = x² + 2ax + a².
6. Simplifying: y² = 4ax.

Diagram: (Visual representation with labeled axes, focus S(a,0), directrix x=-a, and a point P(x,y) showing equal distances PS and PM).

Significance of focus and directrix: The focus determines the 'sharpness' of the parabola, while the directrix acts as a reference line for the definition. Together, they ensure the unique reflective property of parabolas, where any ray parallel to the axis reflects through the focus.

Question 22:
Define a parabola and derive its standard equation in the form y² = 4ax when the vertex is at the origin and the focus lies on the positive x-axis. Include a labeled diagram for clarity.
Answer:

A parabola is the set of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

Derivation of the standard equation y² = 4ax:


1. Let the vertex of the parabola be at the origin (0, 0).
2. Let the focus be at (a, 0), where a > 0 (on the positive x-axis).
3. The directrix is the line x = -a (parallel to the y-axis).
4. Let P(x, y) be any point on the parabola.
5. By definition, the distance of P from the focus = distance of P from the directrix.
6. Using the distance formula:
√[(x - a)² + (y - 0)²] = |x + a|
7. Squaring both sides:
(x - a)² + y² = (x + a)²
8. Expanding both sides:
x² - 2ax + a² + y² = x² + 2ax + a²
9. Simplifying:
y² = 4ax

Diagram: (Visual representation)

  • Draw the x and y axes.
  • Mark the focus at (a, 0).
  • Draw the directrix as a vertical line at x = -a.
  • Sketch the parabola opening to the right, symmetric about the x-axis.

Key Points:

  • The parabola opens rightward because the focus is on the positive x-axis.
  • The latus rectum is the chord passing through the focus and perpendicular to the axis, with length 4a.
  • This standard form is used to model real-world phenomena like satellite dishes and projectile motion.
Question 23:
Derive the standard equation of a parabola with vertex at the origin and focus at (a, 0). Explain each step clearly and state the properties of the parabola.
Answer:

To derive the standard equation of a parabola with vertex at the origin (0, 0) and focus at (a, 0), we follow these steps:


Step 1: Definition of Parabola
A parabola is the locus of a point that moves such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).


Step 2: Assume Directrix
Since the focus is at (a, 0), the directrix will be a vertical line parallel to the y-axis. Let the equation of the directrix be x = -a.


Step 3: Distance Condition
Let P(x, y) be any point on the parabola. According to the definition:
Distance of P from focus = Distance of P from directrix.
√[(x - a)² + (y - 0)²] = |x + a|


Step 4: Squaring Both Sides
(x - a)² + y² = (x + a)²
x² - 2ax + a² + y² = x² + 2ax + a²


Step 5: Simplification
Cancel x² and a² from both sides:
-2ax + y² = 2ax
y² = 4ax


Final Equation: The standard equation of the parabola is y² = 4ax.


Properties of the Parabola:

  • Vertex: (0, 0)
  • Focus: (a, 0)
  • Directrix: x = -a
  • Axis of Symmetry: x-axis (y = 0)
  • Latus Rectum: 4a (length of the chord passing through the focus and perpendicular to the axis)


Application: This form is used in real-world problems like satellite dishes and headlights, where parabolic reflectors are employed to focus signals or light.

Question 24:
Define a parabola and derive its standard equation in the form y² = 4ax. Explain the significance of the focus, directrix, and vertex in the context of a parabola.
Answer:

A parabola is the locus of a point that moves in a plane such that its distance from a fixed point (called the focus) is equal to its distance from a fixed line (called the directrix).

Derivation of the standard equation:
1. Let the focus be at (a, 0) and the directrix be the line x = -a.
2. Consider a point P(x, y) on the parabola.
3. Distance of P from the focus: √[(x - a)² + y²].
4. Distance of P from the directrix: |x + a|.
5. By definition, these distances are equal: √[(x - a)² + y²] = |x + a|.
6. Squaring both sides: (x - a)² + y² = (x + a)².
7. Expanding: x² - 2ax + a² + y² = x² + 2ax + a².
8. Simplifying: y² = 4ax.

Significance of key terms:
- Focus (a, 0): The fixed point that helps define the parabola.
- Directrix (x = -a): The fixed line equidistant from any point on the parabola.
- Vertex (0, 0): The midpoint between the focus and directrix, representing the 'turning point' of the parabola.

This equation represents a parabola opening to the right. The value of a determines the 'width' of the parabola.

Question 25:
Compare and contrast the properties of an ellipse and a hyperbola with respect to their standard equations, eccentricity, and graphical representations. Provide one real-life application for each.
Answer:

Comparison between Ellipse and Hyperbola:

  • Standard Equations:
    - Ellipse: (x²/a²) + (y²/b²) = 1 (horizontal major axis)
    - Hyperbola: (x²/a²) - (y²/b²) = 1 (horizontal transverse axis)
  • Eccentricity (e):
    - Ellipse: 0 ≤ e < 1 (measures 'flatness')
    - Hyperbola: e > 1 (measures 'openness')
  • Graphical Representation:
    - Ellipse: Closed, oval-shaped curve with two foci inside
    - Hyperbola: Open curve with two branches and two foci outside

Real-life Applications:
- Ellipse: Planetary orbits (Kepler's first law states planets move in elliptical orbits with the sun at one focus).
- Hyperbola: Cooling towers of nuclear reactors (their shape provides structural strength and efficient air flow).

Both curves are conic sections but differ fundamentally in their closure and eccentricity ranges. The ellipse represents bounded motion while the hyperbola represents unbounded trajectories.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A parabolic reflector is used in a satellite dish with its focus at (0, 4). The dish has a diameter of 16 meters. Derive the equation of the parabola and calculate its depth at the vertex.
Answer:
Problem Interpretation

We studied that a parabola with vertical axis and vertex at origin has the form x² = 4ay. The focus is given at (0,4), implying a = 4.


Mathematical Modeling

Using the standard form, the equation becomes x² = 16y. The dish's diameter is 16m, so its width is ±8m from the vertex.


Solution

Substituting x = 8 into the equation: 64 = 16y → y = 4. Thus, the depth is 4 meters.

Question 2:
An elliptical arch has a span of 20 meters and height of 8 meters. Find the equation of the ellipse and determine the height at a point 4 meters from the center.
Answer:
Problem Interpretation

Our textbook shows that a horizontal ellipse centered at origin is x²/a² + y²/b² = 1. The span (2a) is 20m → a = 10. The height (b) is 8m.


Mathematical Modeling

The equation becomes x²/100 + y²/64 = 1. We need height (y) when x = 4m.


Solution

Substituting x = 4: 16/100 + y²/64 = 1 → y² = 64(0.84) ≈ 53.76 → y ≈ 7.33m.

Question 3:
A satellite dish is designed as a parabolic reflector with its focus at (0, 2). If the dish is 8 meters wide at the opening, find its depth. Derive the equation of the parabola and calculate the depth.
Answer:
Problem Interpretation

We studied that a parabolic dish has its focus at (0, 2) and a width of 8 meters, implying the parabola opens upwards.


Mathematical Modeling

The standard equation of a parabola with vertex at origin is x² = 4ay. Here, focus (0, a) = (0, 2), so a = 2.


Solution

Substituting a = 2, the equation becomes x² = 8y. At the opening (x = ±4), depth is y = (4)²/8 = 2 meters.

Question 4:
An elliptical arch has a span of 20 meters and a height of 8 meters. Find its equation assuming the major axis is horizontal. Prove that the semi-minor axis is derived from the given dimensions.
Answer:
Problem Interpretation

Our textbook shows that an ellipse with horizontal major axis has equation (x²/a²) + (y²/b²) = 1, where 2a = span and b = height.


Mathematical Modeling

Given span = 20, semi-major axis a = 10. Maximum height occurs at x = 0, so y = b = 8.


Solution

Substituting a = 10 and b = 8, the equation is (x²/100) + (y²/64) = 1. The semi-minor axis b = 8 is directly derived from the arch height.

Question 5:
A parabolic reflector is used in a car headlight. The bulb is placed at the focus of the parabola, and the reflector is 12 cm wide and 4 cm deep.

(i) Find the equation of the parabola.
(ii) How far from the vertex should the bulb be placed?
Answer:
Problem Interpretation

We studied that a parabolic reflector follows the standard equation of a parabola. Given width and depth, we can model it.


Mathematical Modeling

Assume vertex at origin and axis along x-axis. The parabola opens rightward: y² = 4ax.


Solution
  • (i) At depth 4 cm, width is 12 cm ⇒ point (4,6) lies on parabola. Substituting: 6² = 4a(4) ⇒ a = 2.25. Equation: y² = 9x.
  • (ii) Bulb is at focus, a = 2.25 cm from vertex.
Question 6:
An elliptical arch has a span of 40 meters and height of 10 meters at its center.

(i) Derive the standard equation of the ellipse.
(ii) Find the height at a point 8 meters from the center.
Answer:
Problem Interpretation

Our textbook shows that arches often follow elliptical shapes. Given span and height, we can model the ellipse.


Mathematical Modeling

Center the ellipse at origin. Major axis (span) = 40 m ⇒ a = 20. Minor axis (height) = 10 m ⇒ b = 10.


Solution
  • (i) Standard equation: x²/400 + y²/100 = 1.
  • (ii) At x = 8, substitute: 64/400 + y²/100 = 1 ⇒ y = ±6. Height = 6 m.
Question 7:
A parabolic reflector is used in a car headlight. The reflector is 12 cm wide and 6 cm deep. Derive the equation of the parabola and find the position of the bulb if it is placed at the focus.
Answer:
Problem Interpretation

We studied that a parabolic reflector is modeled by a parabola. Given width (12 cm) and depth (6 cm), we derive its equation.

Mathematical Modeling
  • Assume vertex at (0,0) and parabola opening right: y² = 4ax.
  • Point (6,6) lies on it since depth=6 cm and half-width=6 cm.
Solution

Substitute (6,6) into y²=4ax: 36=24a ⇒ a=1.5. Equation: y²=6x. Focus is at (1.5,0), so bulb is 1.5 cm from vertex.

Question 8:
An ellipse represents Earth's orbit with the Sun at one focus. The major axis is 300 million km and eccentricity is 0.0167. Calculate the closest and farthest distances between Earth and Sun.
Answer:
Problem Interpretation

Our textbook shows elliptical orbits follow x²/a² + y²/b² = 1. Given major axis (2a=300) and e=0.0167.

Mathematical Modeling
  • a=150 million km, e=c/a ⇒ c=2.505.
  • Closest (perihelion)=a−c=147.495, farthest (aphelion)=a+c=152.505.
Solution

Distances are 147.495 and 152.505 million km. [Diagram: Ellipse with Sun at focus]

Question 9:
A parabolic reflector is used in a car headlight. The bulb is placed at the focus of the parabola, and the reflector is 20 cm wide and 10 cm deep.

(a) Find the equation of the parabola.
(b) How far from the vertex should the bulb be placed for optimal reflection?
Answer:
Problem Interpretation

We studied that a parabolic reflector follows the standard equation of a parabola. Given its width and depth, we can model it.


Mathematical Modeling

Assume the parabola opens upwards with vertex at (0,0). The width is 20 cm, so endpoints are (±10,10).


Solution
  • (a) Using x² = 4ay, substitute (10,10): 100 = 4a(10) ⇒ a = 2.5. Equation: x² = 10y.
  • (b) The bulb must be placed at the focus (0,a), i.e., 2.5 cm from the vertex.
Question 10:
An elliptical arch has a span of 40 meters and a height of 10 meters at its center.

(a) Derive the standard equation of the ellipse.
(b) Find the height of the arch at a point 8 meters from the center.
Answer:
Problem Interpretation

Our textbook shows that an ellipse's standard form can model arches. Given span (major axis) and height (semi-minor axis), we derive the equation.


Mathematical Modeling

Major axis length = 40 ⇒ 2a = 40 ⇒ a = 20. Semi-minor axis b = 10.


Solution
  • (a) Equation: x²/400 + y²/100 = 1.
  • (b) At x = 8, substitute into the equation: 64/400 + y²/100 = 1 ⇒ y² = 84 ⇒ y ≈ 9.17 meters.
Question 11:
A satellite dish is designed in the shape of a parabola to receive signals. The dish is 2 meters wide and 0.5 meters deep at its center.

(i) Find the equation of the parabola assuming the vertex is at the origin and the parabola opens upwards.

(ii) Determine the position of the receiver (focus) for optimal signal reception.

Answer:

(i) Equation of the parabola:


The parabola has its vertex at the origin (0,0) and opens upwards. Its standard equation is x² = 4ay.
Given the dish is 2 meters wide, the points (1, 0.5) and (-1, 0.5) lie on the parabola (since depth at center is 0.5m).
Substitute (1, 0.5) into the equation:
1² = 4a(0.5)
1 = 2a
a = 0.5
Thus, the equation is x² = 2y.

(ii) Position of the receiver (focus):


For the parabola x² = 4ay, the focus is at (0, a).
Here, a = 0.5, so the focus is at (0, 0.5).
The receiver must be placed 0.5 meters above the vertex for optimal signal reception.
Question 12:
An architect designs an elliptical garden with a major axis of 10 meters and a minor axis of 6 meters.

(i) Write the standard equation of the ellipse if the center is at (0,0) and the major axis is horizontal.

(ii) Calculate the distance between the foci of the ellipse.

Answer:

(i) Standard equation of the ellipse:


Given:
Major axis (2a) = 10m ⇒ a = 5m
Minor axis (2b) = 6m ⇒ b = 3m
The standard equation of an ellipse centered at (0,0) with a horizontal major axis is:
(x²/a²) + (y²/b²) = 1
Substituting values:
(x²/25) + (y²/9) = 1

(ii) Distance between the foci:


The distance between foci is 2c, where c = √(a² - b²).
c = √(25 - 9) = √16 = 4m
Thus, the distance between the foci is 2c = 8m.
Question 13:
A parabolic arch has a span of 20 meters and a height of 10 meters. Find the equation of the parabola assuming the vertex is at the origin and the axis is vertical. Also, determine the height of the arch at a point 4 meters from the center.
Answer:

The given problem involves a parabola with its vertex at the origin and a vertical axis. The standard equation for such a parabola is x² = 4ay, where a is the distance from the vertex to the focus.


Given the span is 20 meters, the parabola passes through the points (10, 10) and (-10, 10) since the height is 10 meters.


Substituting (10, 10) into the equation:
10² = 4a(10)
100 = 40a
a = 2.5


Thus, the equation of the parabola is x² = 10y.


To find the height at 4 meters from the center, substitute x = 4 into the equation:
4² = 10y
16 = 10y
y = 1.6 meters


Therefore, the height of the arch at 4 meters from the center is 1.6 meters.

Question 14:
An ellipse has its major axis along the x-axis with a length of 10 units and minor axis length of 6 units. The center is at (2, -3). Find the standard equation of the ellipse and the coordinates of its foci.
Answer:

The given ellipse has a major axis of 10 units, so 2a = 10 ⇒ a = 5.
The minor axis is 6 units, so 2b = 6 ⇒ b = 3.


The standard equation of an ellipse centered at (h, k) with a horizontal major axis is:
(x - h)²/a² + (y - k)²/b² = 1.


Substituting the given values:
(x - 2)²/25 + (y + 3)²/9 = 1.


To find the foci, we use the relation c² = a² - b², where c is the distance from the center to each focus.
c² = 25 - 9 = 16 ⇒ c = 4.


Since the major axis is horizontal, the foci are located at (h ± c, k).
Thus, the foci are at (2 + 4, -3) and (2 - 4, -3), i.e., (6, -3) and (-2, -3).

Question 15:
A parabolic arch has a span of 20 meters and a height of 10 meters. Find the equation of the parabola assuming its vertex is at the origin and the axis is vertical. Also, determine the height of the arch at a point 4 meters from the center.
Answer:

The given problem describes a parabola with its vertex at the origin and a vertical axis. The standard equation for such a parabola is:

x² = 4ay

Here, the span of the arch is 20 meters, so the parabola passes through the points (10, 10) and (-10, 10) since the height at the ends is 10 meters.

Substituting (10, 10) into the equation:

10² = 4a(10)
100 = 40a
a = 2.5

Thus, the equation of the parabola is:

x² = 10y

To find the height at a point 4 meters from the center, substitute x = 4 into the equation:

4² = 10y
16 = 10y
y = 1.6 meters

Therefore, the height of the arch at 4 meters from the center is 1.6 meters.

Question 16:
An ellipse has its major axis along the x-axis with a length of 10 units and minor axis length of 6 units. The center of the ellipse is at (2, -3). Find the equation of the ellipse and the coordinates of its foci.
Answer:

Given the ellipse has its major axis along the x-axis, the standard form of its equation is:

(x - h)²/a² + (y - k)²/b² = 1

Here, the center (h, k) is (2, -3), the major axis length is 10 units (so a = 5), and the minor axis length is 6 units (so b = 3).

Substituting these values, the equation becomes:

(x - 2)²/25 + (y + 3)²/9 = 1

To find the foci, we use the relationship c² = a² - b²:

c² = 25 - 9 = 16
c = 4

Since the major axis is horizontal, the foci are located at (h ± c, k):

(2 + 4, -3) = (6, -3)
(2 - 4, -3) = (-2, -3)

Thus, the coordinates of the foci are (6, -3) and (-2, -3).

Question 17:
A parabolic arch has a span of 20 meters and a height of 10 meters. Find the equation of the parabola, assuming the vertex is at the origin and the axis is vertical. Also, determine the height of the arch at a horizontal distance of 4 meters from the center.
Answer:

Given the parabolic arch has its vertex at the origin (0, 0) and opens downward, its standard equation is: x² = -4ay.


Step 1: Find the value of a.
At the ends of the arch (x = ±10, y = -10), substitute into the equation:
(10)² = -4a(-10)
100 = 40a
a = 2.5

Step 2: Write the equation.
x² = -4(2.5)y
x² = -10y

Step 3: Find the height at x = 4 meters.
Substitute x = 4 into the equation:
(4)² = -10y
16 = -10y
y = -1.6 meters (below the vertex)

Height from the base = Total height + |y| = 10 + 1.6 = 11.6 meters.
Question 18:
An ellipse has its major axis along the x-axis with a length of 10 units and minor axis length of 6 units. The center is at (2, -3). Find its standard equation and the coordinates of its foci.
Answer:

For an ellipse centered at (h, k) with major axis length 2a and minor axis length 2b, the standard equation is: (x-h)²/a² + (y-k)²/b² = 1.


Step 1: Identify given values.
2a = 10 ⇒ a = 5
2b = 6 ⇒ b = 3
Center (h, k) = (2, -3)

Step 2: Write the equation.
(x-2)²/25 + (y+3)²/9 = 1

Step 3: Find the foci.
Distance of foci from center (c) is given by: c² = a² - b²
c² = 25 - 9 = 16 ⇒ c = 4

Since the major axis is horizontal, foci are at (h ± c, k):
(2 + 4, -3) = (6, -3)
(2 - 4, -3) = (-2, -3)
Question 19:
A satellite dish is designed in the shape of a parabola to receive signals from a satellite. The dish is 2 meters wide and 0.5 meters deep at its center.

(a) Find the equation of the parabola assuming the vertex is at the origin and the parabola opens upwards.

(b) Determine the position of the focus, which is the point where the signals are concentrated.

Answer:

(a) Equation of the parabola:

Given the dish is 2 meters wide and 0.5 meters deep, the parabola has its vertex at the origin (0, 0) and opens upwards. The standard form of such a parabola is x² = 4ay.


At the edges of the dish, the coordinates are (±1, 0.5) because the width is 2 meters (so x = ±1) and depth is 0.5 meters (y = 0.5).


Substitute (1, 0.5) into the equation:
1² = 4a(0.5)
1 = 2a
a = 0.5


Thus, the equation is x² = 2y.


(b) Position of the focus:

For the parabola x² = 4ay, the focus is at (0, a). Here, a = 0.5.


Therefore, the focus is at (0, 0.5) meters, which is the point where signals are concentrated.

Question 20:
An architect designs an elliptical arch for a bridge. The arch has a horizontal span of 10 meters and a maximum height of 4 meters.

(a) Find the standard equation of the ellipse if the center is at the origin.

(b) Calculate the distance between the foci of the ellipse.

Answer:

(a) Standard equation of the ellipse:

Given the horizontal span is 10 meters, the semi-major axis (a) is half of this, so a = 5 meters. The maximum height is 4 meters, so the semi-minor axis (b) is 4 meters.


The standard equation of an ellipse centered at the origin with a horizontal major axis is:
(x²/a²) + (y²/b²) = 1


Substituting a = 5 and b = 4:
(x²/25) + (y²/16) = 1


(b) Distance between the foci:

The distance between the foci is given by 2c, where c = √(a² - b²).


Calculate c:
c = √(25 - 16) = √9 = 3 meters


Thus, the distance between the foci is 6 meters.

Chat on WhatsApp