Conic Sections — Class 11 Maths Notes

Conic Sections · Class 11 Maths · 15 topics.

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Topics covered in Conic Sections

  1. 1.Introduction of Conic Sections

    Conic sections are curves obtained by intersecting a cone with a plane. Depending on the angle of the plane relative to the cone, we get different types of conic sections: circles, ellipses, parabolas, and hyperbolas. These shapes are important in mathematics because they have unique properties and appear frequently in real-world applications.

    Types of Conic Sections

    1. Circle
    2. Ellipse
    3. Parabola
    4. Hyperbola

    Let's explore each type in detail.

    1. Circle

    Definition: A circle is a set of points in a plane that are equidistant from a fixed point called the center.

    Equation: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2

    • (ℎ,𝑘)(h,k) is the center of the circle.
    • 𝑟r is the radius.

    Real-life example: The shape of a wheel or a clock face.

    2. Ellipse

    Definition: An ellipse is a set of points where the sum of the distances from two fixed points (foci) is constant.

    Equation: (𝑥−ℎ)2𝑎2+(𝑦−𝑘)2𝑏2=1a2(x−h)2​+b2(y−k)2​=1

    • (ℎ,𝑘)(h,k) is the center.
    • 𝑎a and 𝑏b are the semi-major and semi-minor axes.

    Real-life example: The orbits of planets around the sun.

    3. Parabola

    Definition: A parabola is a set of points where each point is equidistant from a fixed point (focus) and a fixed line (directrix).

    Equation: 𝑦=𝑎𝑥2+𝑏𝑥+𝑐y=ax2+bx+c (standard form)

    • The vertex is the turning point of the parabola.

    Real-life example: The shape of satellite dishes or the path of a thrown ball.

    4. Hyperbola

    Definition: A hyperbola is a set of points where the difference of the distances from two fixed points (foci) is constant.

    Equation: (𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1a2(x−h)2​−b2(y−k)2​=1

    • (ℎ,𝑘)(h,k) is the center.
    • 𝑎a and 𝑏b are distances that define the shape.

    Real-life example: The shape of cooling towers of nuclear power plants.

    Real-World Applications

    • Engineering: Parabolic reflectors, satellite dishes.
    • Astronomy: Orbits of planets and comets (ellipses).
    • Physics: Path of projectiles (parabolas).
    • Architecture: Design of structures like arches and bridges (parabolas and hyperbolas).

    Careers Involving Conic Sections

    • Astronomy: Understanding planetary orbits.
    • Engineering: Designing optical systems and structures.
    • Physics: Analyzing motion and forces.
    • Architecture: Designing aesthetically pleasing and structurally sound buildings.

  2. 2.Sections of a Cone

    Conic sections are the curves obtained by intersecting a right circular cone with a plane. The shape of the conic section depends on the angle at which the plane intersects the cone. Let's explore the different sections in detail.

    Types of Conic Sections

    1. Circle
    2. Ellipse
    3. Parabola
    4. Hyperbola

    We will understand each of these conic sections by examining how they are formed.

    1. Circle

    Formation: When a plane cuts the cone parallel to its base, the intersection is a circle. This happens because every point on the plane is equidistant from the cone's axis.

    Properties:

    • All points are equidistant from the center.
    • Equation: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2
      • (ℎ,𝑘)(h,k) is the center.
      • 𝑟r is the radius.

    Real-life example: The face of a clock or the shape of a coin.

    2. Ellipse

    Formation: When a plane cuts through the cone at an angle, but not parallel to the base and not passing through the apex, the intersection is an ellipse.

    Properties:

    • Sum of the distances from any point on the ellipse to two fixed points (foci) is constant.
    • Equation: (𝑥−ℎ)2𝑎2+(𝑦−𝑘)2𝑏2=1a2(x−h)2​+b2(y−k)2​=1
      • (ℎ,𝑘)(h,k) is the center.
      • 𝑎a and 𝑏b are the lengths of the semi-major and semi-minor axes.

    Real-life example: The orbit of planets around the sun.

    3. Parabola

    Formation: When a plane is parallel to one of the slant heights of the cone, the intersection is a parabola.

    Properties:

    • Each point on a parabola is equidistant from a fixed point (focus) and a fixed line (directrix).
    • Equation: 𝑦=𝑎𝑥2+𝑏𝑥+𝑐y=ax2+bx+c
      • The vertex is the point where the parabola changes direction.

    Real-life example: The path of a projectile like a ball thrown in the air, or the shape of satellite dishes.

    4. Hyperbola

    Formation: When a plane cuts both nappes (the upper and lower parts) of the cone, the intersection is a hyperbola.

    Properties:

    • Difference of the distances from any point on the hyperbola to two fixed points (foci) is constant.
    • Equation: (𝑥−ℎ)2𝑎2−(𝑦−𝑘)2𝑏2=1a2(x−h)2​−b2(y−k)2​=1
      • (ℎ,𝑘)(h,k) is the center.
      • 𝑎a and 𝑏b are distances that define the shape.

    Real-life example: The shape of cooling towers in power plants.

    Visualization of Conic Sections

    To better understand, visualize a right circular cone. Imagine slicing it with different planes:

    • Parallel to the base: You get a circle.
    • At an angle (not parallel to the base and not through the apex): You get an ellipse.
    • Parallel to a slant height: You get a parabola.
    • Through both nappes: You get a hyperbola.

    Real-World Applications

    1. Circles: Wheels, clocks, coins.
    2. Ellipses: Orbits of celestial bodies, whispering galleries.
    3. Parabolas: Satellite dishes, car headlights, projectile motion.
    4. Hyperbolas: Radio and radar signals, cooling towers.

    Careers Using Conic Sections

    1. Astronomy: Studying planetary orbits.
    2. Engineering: Designing optical systems like telescopes and satellites.
    3. Physics: Analyzing motion and forces.
    4. Architecture: Designing arches, bridges, and other structures.

  3. 3.Degenerated Conic Sections

    When we talk about conic sections, we usually refer to the standard shapes: circles, ellipses, parabolas, and hyperbolas. However, under certain conditions, the intersection of a cone and a plane can result in what are called degenerated conic sections. These are special cases where the conic section collapses into simpler geometric shapes.

    Types of Degenerated Conic Sections

    1. Point
    2. Line
    3. Pair of Intersecting Lines

    Let's explore each type and understand how they are formed.

    1. Point

    Formation: When the intersecting plane passes through the apex (vertex) of the cone, the conic section degenerates into a single point. This point is the apex itself.

    Equation: If the equation of the conic section reduces to 0=00=0, it signifies a point.

    Example: This occurs when the plane touches the cone at exactly one point.

    Visual Representation:

    • Imagine a plane slicing through the very tip of the cone; the result is just a point.

    2. Line

    Formation: When the plane is tangent to the cone, intersecting it in such a way that it just touches the cone along a line, the conic section degenerates into a single straight line.

    Equation: If the conic section's equation can be factored into two identical linear factors, it signifies a line.

    Example: This occurs when the plane is parallel to one of the generating lines of the cone.

    Visual Representation:

    • Imagine a plane that just grazes the side of the cone, touching it along a straight line from the apex to the base.

    3. Pair of Intersecting Lines

    Formation: When the plane passes through the apex of the cone and intersects both nappes (the two parts of the cone), the conic section degenerates into a pair of intersecting lines. These lines intersect at the apex.

    Equation: If the equation of the conic section can be factored into two distinct linear factors, it signifies a pair of intersecting lines.

    Example: This occurs when the plane cuts through the cone's apex and at an angle, intersecting both the upper and lower parts of the cone.

    Visual Representation:

    • Imagine a plane that cuts through the very center of the cone and continues to cut through both parts, forming an "X" shape.

    Real-World Examples and Applications

    • Point: Can be seen in the precision required in fields like engineering and manufacturing, where exact points are crucial.
    • Line: Tangent lines are used in calculus to determine slopes and rates of change. In optics, they represent the path of light rays.
    • Pair of Intersecting Lines: These are seen in structural designs and architectural elements, where intersecting lines can represent supports or beams.

    Careers Using Degenerated Conic Sections

    1. Engineering: Understanding points and lines is fundamental in designing machinery and structures.
    2. Physics: Analysis of light paths and forces often involves tangents and intersections.
    3. Architecture: Designing and analyzing structures that may involve intersecting lines for support.
    4. Mathematics: The study of geometry and algebra relies on understanding these fundamental shapes.

  4. 4.Circle

    Definition of a Circle

    A circle is a set of all points in a plane that are at a fixed distance (called the radius) from a fixed point (called the center).

    Key Terms:

    • Center (O or C): The fixed point from which every point on the circle is equidistant.
    • Radius (r): The constant distance from the center to any point on the circle.
    • Diameter: A chord that passes through the center of the circle, which is twice the radius.

    Diagram Explanation

    Let's go through each part of the provided diagrams to understand the circle better.

    1. Left Diagram: Basic Circle Properties

      • Center (O): The point O is the center of the circle.
      • Radius (OP1, OP2, OP3): The lengths OP1, OP2, and OP3 are equal, representing the radius of the circle.
      • Points on the Circle (P1, P2, P3): These are points on the circumference of the circle, equidistant from the center.

    2. Right Diagram: Circle in Coordinate Plane

      • Center (C): The center of the circle is at the point (h, k) in the coordinate plane.
      • Point on the Circle (P): Any point on the circle can be represented as (x, y).
      • The radius is the distance between the center (C) and any point (P) on the circle.

    Equation of a Circle

    In the coordinate plane, the equation of a circle with center at (h, k) and radius r is given by: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2

    Example in Real Life

    Imagine a round park with a fountain at the center. If you walk around the edge of the park, the distance from the fountain (the center) to any point on your path (the circumference) remains the same. This distance is the radius of the circle that represents the park.

    Step-by-Step Derivation

    1. Identify the center (h, k) and a point on the circle (x, y).
    2. Calculate the distance between (h, k) and (x, y) using the distance formula: Distance=(𝑥−ℎ)2+(𝑦−𝑘)2Distance=(x−h)2+(y−k)2​
    3. Set this distance equal to the radius (r): (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟(x−h)2+(y−k)2​=r
    4. Square both sides to eliminate the square root: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2

    This is the standard form of the equation of a circle.

  5. 5.Exercise Questions

    Exercise 1

    Find the equation of the circle with center (0, 2) and radius 2.

    Solution: The equation of a circle with center (ℎ,𝑘)(h,k) and radius 𝑟r is given by: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2

    Here, the center is (0,2)(0,2) and the radius is 22: (𝑥−0)2+(𝑦−2)2=22(x−0)2+(y−2)2=22 𝑥2+(𝑦−2)2=4x2+(y−2)2=4

    Exercise 2

    Find the equation of the circle with center (2,3)(2,3) and radius 4.

    Solution: The center is (2,3)(2,3) and the radius is 44: (𝑥−2)2+(𝑦−3)2=42(x−2)2+(y−3)2=42 (𝑥−2)2+(𝑦−3)2=16(x−2)2+(y−3)2=16

    Exercise 3

    Find the equation of the circle with center (12,14)(21​,41​) and radius 112121​.

    Solution: The center is (12,14)(21​,41​) and the radius is 112121​: (𝑥−12)2+(𝑦−14)2=(112)2(x−21​)2+(y−41​)2=(121​)2 (𝑥−12)2+(𝑦−14)2=1144(x−21​)2+(y−41​)2=1441​

    Exercise 4

    Find the equation of the circle with center (1,1)(1,1) and radius 22​.

    Solution: The center is (1,1)(1,1) and the radius is 22​: (𝑥−1)2+(𝑦−1)2=(2)2(x−1)2+(y−1)2=(2​)2 (𝑥−1)2+(𝑦−1)2=2(x−1)2+(y−1)2=2

    Exercise 5

    Find the equation of the circle with center (−𝑎,−𝑏)(−a,−b) and radius 𝑎2−𝑏2a2−b2​.

    Solution: The center is (−𝑎,−𝑏)(−a,−b) and the radius is 𝑎2−𝑏2a2−b2​: (𝑥+𝑎)2+(𝑦+𝑏)2=(𝑎2−𝑏2)2(x+a)2+(y+b)2=(a2−b2​)2 (𝑥+𝑎)2+(𝑦+𝑏)2=𝑎2−𝑏2(x+a)2+(y+b)2=a2−b2

    Exercise 6

    Find the center and radius of the circle given by the equation: (𝑥+5)2+(𝑦−3)2=36(x+5)2+(y−3)2=36

    Solution: The equation is already in the standard form (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2:

    • Center: (−5,3)(−5,3)
    • Radius: 36=636​=6

    Exercise 7

    Find the center and radius of the circle given by the equation: 𝑥2+𝑦2−4𝑥−8𝑦−45=0x2+y2−4x−8y−45=0

    Solution: Complete the square to transform the equation into the standard form: 𝑥2−4𝑥+𝑦2−8𝑦=45x2−4x+y2−8y=45 (𝑥2−4𝑥+4)+(𝑦2−8𝑦+16)=45+4+16(x2−4x+4)+(y2−8y+16)=45+4+16 (𝑥−2)2+(𝑦−4)2=65(x−2)2+(y−4)2=65

    • Center: (2,4)(2,4)
    • Radius: 6565​

    Exercise 8

    Find the center and radius of the circle given by the equation: 𝑥2+𝑦2−8𝑥+10𝑦−12=0x2+y2−8x+10y−12=0

    Solution: Complete the square: 𝑥2−8𝑥+𝑦2+10𝑦=12x2−8x+y2+10y=12 (𝑥2−8𝑥+16)+(𝑦2+10𝑦+25)=12+16+25(x2−8x+16)+(y2+10y+25)=12+16+25 (𝑥−4)2+(𝑦+5)2=53(x−4)2+(y+5)2=53

    • Center: (4,−5)(4,−5)
    • Radius: 5353​

    Exercise 9

    Find the center and radius of the circle given by the equation: 2𝑥2+2𝑦2−𝑥=02x2+2y2−x=0

    Solution: Divide the entire equation by 2: 𝑥2+𝑦2−12𝑥=0x2+y2−21​x=0 Complete the square: 𝑥2−12𝑥+𝑦2=0x2−21​x+y2=0 (𝑥2−12𝑥+116)+𝑦2=116(x2−21​x+161​)+y2=161​ (𝑥−14)2+𝑦2=116(x−41​)2+y2=161​

    • Center: (14,0)(41​,0)
    • Radius: 1441​

    Exercise 10

    Find the equation of the circle passing through the points (4,1)(4,1) and (6,5)(6,5) and whose center is on the line 4𝑥+𝑦=164x+y=16.

    Solution:

    1. Let the center be (ℎ,𝑘)(h,k).
    2. Since the circle passes through (4,1)(4,1) and (6,5)(6,5):(ℎ−4)2+(𝑘−1)2=(ℎ−6)2+(𝑘−5)2(h−4)2+(k−1)2​=(h−6)2+(k−5)2​
    3. Also, the center lies on the line:4ℎ+𝑘=164h+k=16

    We have two conditions:

    {(ℎ−4)2+(𝑘−1)2=(ℎ−6)2+(𝑘−5)24ℎ+𝑘=16{(h−4)2+(k−1)2​=(h−6)2+(k−5)2​4h+k=16​

    1. Solving for the first equation:

      (ℎ−4)2+(𝑘−1)2=(ℎ−6)2+(𝑘−5)2(h−4)2+(k−1)2=(h−6)2+(k−5)2ℎ2−8ℎ+16+𝑘2−2𝑘+1=ℎ2−12ℎ+36+𝑘2−10𝑘+25h2−8h+16+k2−2k+1=h2−12h+36+k2−10k+25−8ℎ+17=−12ℎ+61−8h+17=−12h+614ℎ=444h=44ℎ=11h=11

    2. Substitute ℎ=11h=11 into 4ℎ+𝑘=164h+k=16:

      4(11)+𝑘=164(11)+k=1644+𝑘=1644+k=16𝑘=16−44k=16−44𝑘=−28k=−28

    3. The center of the circle is (11,−28)(11,−28).

    4. Radius calculation using the point (4,1)(4,1):

      𝑟=(11−4)2+(−28−1)2r=(11−4)2+(−28−1)2​𝑟=72+(−29)2r=72+(−29)2​𝑟=49+841r=49+841​𝑟=890r=890​𝑟=890r=890​

    5. Equation of the circle:

      (𝑥−11)2+(𝑦+28)2=890(x−11)2+(y+28)2=890

  6. 6.Exercise Questions

    Question 1

    Find the equation of the circle passing through (0,0)(0,0) and making intercepts 𝑎a and 𝑏b on the coordinate axes.

    Solution:

    1. The general equation of a circle is: 𝑥2+𝑦2+2𝑔𝑥+2𝑓𝑦+𝑐=0x2+y2+2gx+2fy+c=0

    2. Since the circle passes through the origin (0,0)(0,0), substitute 𝑥=0x=0 and 𝑦=0y=0: 0+0+2𝑔(0)+2𝑓(0)+𝑐=00+0+2g(0)+2f(0)+c=0 𝑐=0c=0

    3. Therefore, the equation reduces to: 𝑥2+𝑦2+2𝑔𝑥+2𝑓𝑦=0x2+y2+2gx+2fy=0

    4. Since the circle makes intercepts 𝑎a and 𝑏b on the coordinate axes, it passes through (𝑎,0)(a,0) and (0,𝑏)(0,b):

      • Substitute (𝑎,0)(a,0): 𝑎2+2𝑔𝑎=0a2+2ga=0 𝑔=−𝑎2g=−2a​

      • Substitute (0,𝑏)(0,b): 𝑏2+2𝑓𝑏=0b2+2fb=0 𝑓=−𝑏2f=−2b​

    5. Now, substitute 𝑔g and 𝑓f back into the equation: 𝑥2+𝑦2−𝑎𝑥−𝑏𝑦=0x2+y2−ax−by=0

    Question 2

    Find the equation of a circle with center (2,2)(2,2) and passing through the point (4,5)(4,5).

    Solution:

    1. The general equation of a circle with center (ℎ,𝑘)(h,k) and radius 𝑟r is: (𝑥−ℎ)2+(𝑦−𝑘)2=𝑟2(x−h)2+(y−k)2=r2

    2. Here, the center is (2,2)(2,2), so: (𝑥−2)2+(𝑦−2)2=𝑟2(x−2)2+(y−2)2=r2

    3. To find the radius, use the given point (4,5)(4,5): 𝑟=(4−2)2+(5−2)2r=(4−2)2+(5−2)2​ 𝑟=22+32r=22+32​ 𝑟=4+9r=4+9​ 𝑟=13r=13​

    4. Substitute 𝑟r back into the equation: (𝑥−2)2+(𝑦−2)2=13(x−2)2+(y−2)2=13

    Question 3

    Does the point (−2.5,3.5)(−2.5,3.5) lie inside, outside, or on the circle 𝑥2+𝑦2=25x2+y2=25?

    Solution:

    1. First, determine the distance from the point (−2.5,3.5)(−2.5,3.5) to the origin (center of the circle): Distance=(−2.5)2+(3.5)2Distance=(−2.5)2+(3.5)2​ Distance=6.25+12.25Distance=6.25+12.25​ Distance=18.5Distance=18.5​ Distance≈4.3Distance≈4.3

    2. Compare this distance with the radius of the circle:

      • The radius of the circle 𝑥2+𝑦2=25x2+y2=25 is 25=525​=5.

    3. Since 4.3<54.3<5, the point (−2.5,3.5)(−2.5,3.5) lies inside the circle.

  7. 7.Parabola

    A parabola is a U-shaped curve that can open upwards, downwards, left, or right. It is defined as 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.

    Key Terms:

    • Focus (F): The fixed point inside the parabola.
    • Directrix: The fixed line outside the parabola.
    • Vertex (V): The midpoint between the focus and the directrix; it is the point where the parabola changes direction.
    • Axis of Symmetry: The line that passes through the focus and the vertex, dividing the parabola into two mirror-image halves.
    • Latus Rectum: The line segment perpendicular to the axis of symmetry and passing through the focus, with its endpoints on the parabola.

    Standard Equations of Parabola

    The standard form of a parabola's equation depends on its orientation. Here are the different orientations and their standard equations:

    1. Parabola Opening Upwards or Downwards

    Equation: 𝑦2=4𝑎𝑥y2=4ax

    • Vertex: (0,0)(0,0)
    • Focus: (𝑎,0)(a,0)
    • Directrix: 𝑥=−𝑎x=−a
    • Axis of Symmetry: 𝑦=0y=0

    2. Parabola Opening Left or Right

    Equation: 𝑥2=4𝑎𝑦x2=4ay

    • Vertex: (0,0)(0,0)
    • Focus: (0,𝑎)(0,a)
    • Directrix: 𝑦=−𝑎y=−a
    • Axis of Symmetry: 𝑥=0x=0

    General Form with Vertex at (ℎ,𝑘)(h,k)

    When the vertex is not at the origin, but at (ℎ,𝑘)(h,k), the equations are adjusted as follows:

    3. Parabola Opening Upwards

    Equation: (𝑥−ℎ)2=4𝑎(𝑦−𝑘)(x−h)2=4a(y−k)

    • Vertex: (ℎ,𝑘)(h,k)
    • Focus: (ℎ,𝑘+𝑎)(h,k+a)
    • Directrix: 𝑦=𝑘−𝑎y=k−a
    • Axis of Symmetry: 𝑥=ℎx=h

    4. Parabola Opening Downwards

    Equation: (𝑥−ℎ)2=−4𝑎(𝑦−𝑘)(x−h)2=−4a(y−k)

    • Vertex: (ℎ,𝑘)(h,k)
    • Focus: (ℎ,𝑘−𝑎)(h,k−a)
    • Directrix: 𝑦=𝑘+𝑎y=k+a
    • Axis of Symmetry: 𝑥=ℎx=h

    5. Parabola Opening Right

    Equation: (𝑦−𝑘)2=4𝑎(𝑥−ℎ)(y−k)2=4a(x−h)

    • Vertex: (ℎ,𝑘)(h,k)
    • Focus: (ℎ+𝑎,𝑘)(h+a,k)
    • Directrix: 𝑥=ℎ−𝑎x=h−a
    • Axis of Symmetry: 𝑦=𝑘y=k

    6. Parabola Opening Left

    Equation: (𝑦−𝑘)2=−4𝑎(𝑥−ℎ)(y−k)2=−4a(x−h)

    • Vertex: (ℎ,𝑘)(h,k)
    • Focus: (ℎ−𝑎,𝑘)(h−a,k)
    • Directrix: 𝑥=ℎ+𝑎x=h+a
    • Axis of Symmetry: 𝑦=𝑘y=k

    Real-life Example

    Imagine you are holding a flashlight. The shape of the light beam that spreads out in front of you is a parabolic shape. This is because the reflector inside the flashlight is parabolic, directing the light rays outwards in a focused manner.

  8. 8.Latus rectum

    The latus rectum of a parabola is a line segment that is perpendicular to the axis of symmetry of the parabola, passes through the focus, and has its endpoints on the parabola. The length of the latus rectum is a measure of how "wide" the parabola is at the level of the focus.

    Key Points:

    • The latus rectum is parallel to the directrix.
    • Its length is equal to 4𝑎4a, where 𝑎a is the distance from the vertex to the focus of the parabola.

    Standard Equations and Latus Rectum:

    1. For the parabola 𝑦2=4𝑎𝑥y2=4ax:

      • Focus: (𝑎,0)(a,0)
      • Length of latus rectum: 4𝑎4a
      • Endpoints of the latus rectum: (𝑎,2𝑎)(a,2a) and (𝑎,−2𝑎)(a,−2a)

    2. For the parabola 𝑥2=4𝑎𝑦x2=4ay:

      • Focus: (0,𝑎)(0,a)
      • Length of latus rectum: 4𝑎4a
      • Endpoints of the latus rectum: (2𝑎,𝑎)(2a,a) and (−2𝑎,𝑎)(−2a,a)

    Problem-Solving Questions:

    Question 1

    Find the length of the latus rectum and the coordinates of its endpoints for the parabola 𝑦2=12𝑥y2=12x.

    Solution:

    1. Identify the given equation: 𝑦2=12𝑥y2=12x

    2. Compare it with the standard form 𝑦2=4𝑎𝑥y2=4ax to find 𝑎a: 4𝑎=124a=12 𝑎=3a=3

    3. Length of the latus rectum: Length=4𝑎=4×3=12Length=4a=4×3=12

    4. Endpoints of the latus rectum:

      • The focus is at (𝑎,0)=(3,0)(a,0)=(3,0).
      • The endpoints are (3,2𝑎)=(3,6)(3,2a)=(3,6) and (3,−2𝑎)=(3,−6)(3,−2a)=(3,−6).

    So, the length of the latus rectum is 12, and the endpoints are (3,6)(3,6) and (3,−6)(3,−6).

    Question 2

    Find the length of the latus rectum and the coordinates of its endpoints for the parabola 𝑥2=−8𝑦x2=−8y.

    Solution:

    1. Identify the given equation: 𝑥2=−8𝑦x2=−8y

    2. Compare it with the standard form 𝑥2=4𝑎𝑦x2=4ay to find 𝑎a: 4𝑎=−84a=−8 𝑎=−2a=−2

    3. Length of the latus rectum: Length=4∣𝑎∣=4×2=8Length=4∣a∣=4×2=8

    4. Endpoints of the latus rectum:

      • The focus is at (0,𝑎)=(0,−2)(0,a)=(0,−2).
      • The endpoints are (2𝑎,−2)=(4,−2)(2a,−2)=(4,−2) and (−2𝑎,−2)=(−4,−2)(−2a,−2)=(−4,−2).

    So, the length of the latus rectum is 8, and the endpoints are (4,−2)(4,−2) and (−4,−2)(−4,−2).

  9. 9.Exercise Questions

    For each of the following exercises, find the coordinates of the focus, the axis of the parabola, the equation of the directrix, and the length of the latus rectum.

    Exercise 1

    𝑦2=12𝑥y2=12x

    1. Equation in standard form: 𝑦2=4𝑎𝑥y2=4ax

      • 4𝑎=124a=12 so, 𝑎=3a=3

    2. Coordinates of the focus: (𝑎,0)=(3,0)(a,0)=(3,0)

    3. Axis of the parabola: 𝑦=0y=0 (horizontal axis)

    4. Equation of the directrix: 𝑥=−𝑎x=−a

      • 𝑥=−3x=−3

    5. Length of the latus rectum: 4𝑎=124a=12

    Exercise 2

    𝑥2=6𝑦x2=6y

    1. Equation in standard form: 𝑥2=4𝑎𝑦x2=4ay

      • 4𝑎=64a=6 so, 𝑎=32a=23​

    2. Coordinates of the focus: (0,𝑎)=(0,32)(0,a)=(0,23​)

    3. Axis of the parabola: 𝑥=0x=0 (vertical axis)

    4. Equation of the directrix: 𝑦=−𝑎y=−a

      • 𝑦=−32y=−23​

    5. Length of the latus rectum: 4𝑎=64a=6

    Exercise 3

    𝑦2=−8𝑥y2=−8x

    1. Equation in standard form: 𝑦2=4𝑎𝑥y2=4ax

      • 4𝑎=−84a=−8 so, 𝑎=−2a=−2

    2. Coordinates of the focus: (𝑎,0)=(−2,0)(a,0)=(−2,0)

    3. Axis of the parabola: 𝑦=0y=0 (horizontal axis)

    4. Equation of the directrix: 𝑥=−𝑎x=−a

      • 𝑥=2x=2

    5. Length of the latus rectum: 4∣𝑎∣=84∣a∣=8

    Exercise 4

    𝑥2=−16𝑦x2=−16y

    1. Equation in standard form: 𝑥2=4𝑎𝑦x2=4ay

      • 4𝑎=−164a=−16 so, 𝑎=−4a=−4

    2. Coordinates of the focus: (0,𝑎)=(0,−4)(0,a)=(0,−4)

    3. Axis of the parabola: 𝑥=0x=0 (vertical axis)

    4. Equation of the directrix: 𝑦=−𝑎y=−a

      • 𝑦=4y=4

    5. Length of the latus rectum: 4∣𝑎∣=164∣a∣=16

    Exercise 5

    𝑦2=10𝑥y2=10x

    1. Equation in standard form: 𝑦2=4𝑎𝑥y2=4ax

      • 4𝑎=104a=10 so, 𝑎=52a=25​

    2. Coordinates of the focus: (𝑎,0)=(52,0)(a,0)=(25​,0)

    3. Axis of the parabola: 𝑦=0y=0 (horizontal axis)

    4. Equation of the directrix: 𝑥=−𝑎x=−a

      • 𝑥=−52x=−25​

    5. Length of the latus rectum: 4𝑎=104a=10

    Exercise 6

    𝑥2=−9𝑦x2=−9y

    1. Equation in standard form: 𝑥2=4𝑎𝑦x2=4ay

      • 4𝑎=−94a=−9 so, 𝑎=−94a=−49​

    2. Coordinates of the focus: (0,𝑎)=(0,−94)(0,a)=(0,−49​)

    3. Axis of the parabola: 𝑥=0x=0 (vertical axis)

    4. Equation of the directrix: 𝑦=−𝑎y=−a

      • 𝑦=94y=49​

    5. Length of the latus rectum: 4∣𝑎∣=94∣a∣=9

    Exercises 7 to 12

    Find the equation of the parabola that satisfies the given conditions:

    Exercise 7

    Focus (6,0); directrix 𝑥=−6x=−6

    1. The distance from the vertex to the focus is 6.
    2. The distance from the vertex to the directrix is also 6.
    3. The vertex is the midpoint of the focus and the directrix, so the vertex is at the origin (0, 0).
    4. Equation of the parabola opening right is 𝑦2=4𝑎𝑥y2=4ax.
    5. Here, 4𝑎=244a=24, so 𝑎=6a=6.

    𝑦2=24𝑥y2=24x

    Exercise 8

    Focus (0,-3); directrix 𝑦=3y=3

    1. The distance from the vertex to the focus is 3.
    2. The distance from the vertex to the directrix is also 3.
    3. The vertex is the midpoint of the focus and the directrix, so the vertex is at the origin (0, 0).
    4. Equation of the parabola opening down is 𝑥2=−4𝑎𝑦x2=−4ay.
    5. Here, 4𝑎=124a=12, so 𝑎=3a=3.

    𝑥2=−12𝑦x2=−12y

    Exercise 9

    Vertex (0,0); focus (3,0)

    1. The distance from the vertex to the focus is 3.
    2. Equation of the parabola opening right is 𝑦2=4𝑎𝑥y2=4ax.
    3. Here, 𝑎=3a=3, so 4𝑎=124a=12.

    𝑦2=12𝑥y2=12x

    Exercise 10

    Vertex (0,0); focus (0,-2)

    1. The distance from the vertex to the focus is 2.
    2. Equation of the parabola opening down is 𝑥2=−4𝑎𝑦x2=−4ay.
    3. Here, 𝑎=2a=2, so 4𝑎=84a=8.

    𝑥2=−8𝑦x2=−8y

    Exercise 11

    Vertex (0,0) passing through (2,3) and axis is along x-axis.

    1. The equation of the parabola opening right is 𝑦2=4𝑎𝑥y2=4ax.
    2. Substitute point (2, 3) into the equation to find 𝑎a:

    32=4𝑎⋅232=4a⋅2 9=8𝑎9=8a 𝑎=98a=89​

    𝑦2=92𝑥y2=29​x

    Exercise 12

    Vertex (0,0) passing through (5,2) and symmetric with respect to y-axis.

    1. The equation of the parabola opening up is 𝑥2=4𝑎𝑦x2=4ay.
    2. Substitute point (5, 2) into the equation to find 𝑎a:

    52=4𝑎⋅252=4a⋅2 25=8𝑎25=8a 𝑎=258a=825​

    𝑥2=252𝑦x2=225​y

  10. 10.Ellipse

    1. Definition and Basic Properties

    Definition: An ellipse is a set of points in a plane such that the sum of the distances from any point on the ellipse to two fixed points (called foci) is constant.

    Basic Properties:

    • Major Axis: The longest diameter of the ellipse, passing through both foci.
    • Minor Axis: The shortest diameter of the ellipse, perpendicular to the major axis at the center.
    • Center: The midpoint of the major and minor axes.
    • Foci (singular: focus): Two fixed points on the major axis, equidistant from the center.
    • Vertices: The endpoints of the major axis.

    2. Equation of an Ellipse

    The standard form of the equation of an ellipse with the center at the origin (0,0)(0,0) is: 𝑥2𝑎2+𝑦2𝑏2=1a2x2​+b2y2​=1

    Where:

    • 𝑎a is the semi-major axis length.
    • 𝑏b is the semi-minor axis length.

    If 𝑎>𝑏a>b, the ellipse is elongated along the x-axis. If 𝑏>𝑎b>a, the ellipse is elongated along the y-axis.

    3. Deriving the Equation

    To derive the equation, let's consider an ellipse centered at the origin with foci at (𝑐,0)(c,0) and (−𝑐,0)(−c,0).

    The distance from a point (𝑥,𝑦)(x,y) on the ellipse to the foci (𝑐,0)(c,0) and (−𝑐,0)(−c,0) is given by:

    (𝑥−𝑐)2+𝑦2+(𝑥+𝑐)2+𝑦2=2𝑎(x−c)2+y2​+(x+c)2+y2​=2a

    Here, 2𝑎2a is the constant sum of the distances.

    Using algebra and simplifying, we arrive at the standard form of the ellipse equation:

    𝑥2𝑎2+𝑦2𝑏2=1a2x2​+b2y2​=1

    where 𝑏2=𝑎2−𝑐2b2=a2−c2.

    4. Real-Life Example

    Imagine you are standing in an oval-shaped running track. The two foci of the ellipse could be marked on the track. No matter where you stand on the track, the sum of your distances to these two fixed points will always be the same.

    5. Applications of Ellipses

    Ellipses have many practical applications, including:

    • Astronomy: The orbits of planets and satellites are often elliptical.
    • Engineering: Elliptical gears and arches are used for smooth mechanical motion and strong structural designs.
    • Medicine: Elliptical machines are used in cardiovascular exercises.

    6. Careers and Industries

    • Astronomy: Studying celestial bodies and their orbits.
    • Engineering: Designing efficient mechanical systems.
    • Architecture: Creating aesthetically pleasing and structurally sound buildings.

    Activities to Understand Better

    1. Drawing an Ellipse: Use two pins, a string, and a pencil. Place the pins at the foci, loop the string around them, and use the pencil to trace the ellipse.
    2. Finding the Foci: Given the lengths of the major and minor axes, calculate the distance to the foci using 𝑐=𝑎2−𝑏2c=a2−b2​.

  11. 11.Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse.

    Let's analyze the relationship between the semi-major axis, semi-minor axis, and the distance of the focus from the center of the ellipse using the provided diagram.

    Key Elements in the Diagram

    • O: Center of the ellipse.
    • F1 and F2: Foci of the ellipse.
    • P and R: Points on the major axis (vertices of the ellipse).
    • Q: Point on the ellipse, not on the axes.
    • a: Semi-major axis length.
    • b: Semi-minor axis length.
    • c: Distance from the center to each focus (F1 and F2).

    Relationships and Formulas

    1. Distance from Center to a Focus (𝑐c):

      𝑐=𝑎2−𝑏2c=a2−b2​

    2. Sum of Distances from Any Point on the Ellipse to the Foci:

      𝑃𝐹1+𝑃𝐹2=2𝑎PF1​+PF2​=2a

      This is the defining property of an ellipse.

    3. Length of Major Axis (Total):

      2𝑎2a

    4. Length of Minor Axis (Total):

      2𝑏2b

    Explanation Using the Diagram

    • The center 𝑂O is the midpoint of the major and minor axes.
    • The distance 𝑎a is the length from the center 𝑂O to a vertex 𝑃P on the major axis.
    • The distance 𝑏b is the length from the center 𝑂O to a point 𝑄Q on the ellipse along the minor axis.
    • The foci 𝐹1F1 and 𝐹2F2 are located symmetrically along the major axis, each at a distance 𝑐c from the center 𝑂O.

    Proof of the Relationship 𝑐=𝑎2−𝑏2c=a2−b2​

    Consider the right triangle formed by the points 𝑂O (center), 𝑄Q (a point on the ellipse along the minor axis), and 𝑃P (vertex along the major axis):

    • The hypotenuse is the semi-major axis 𝑎a.
    • One leg is the semi-minor axis 𝑏b.
    • The other leg is the distance from the center to the focus 𝑐c.

    Using the Pythagorean theorem:

    𝑎2=𝑏2+𝑐2a2=b2+c2

    Solving for 𝑐c:

    𝑐=𝑎2−𝑏2c=a2−b2​

    This equation shows how the distance of the foci from the center is related to the lengths of the semi-major and semi-minor axes.

    Example in Real Life

    Consider an elliptical running track where the length from the center to the edge along the longest side is 100 meters (semi-major axis), and the length from the center to the edge along the shortest side is 60 meters (semi-minor axis). The foci will be at a distance of:

    𝑐=1002−602=10000−3600=6400=80 metersc=1002−602​=10000−3600​=6400​=80 meters

    This means the two foci are each 80 meters from the center of the track.

    Careers and Industries

    Understanding ellipses is crucial in fields like:

    • Astronomy: For calculating planetary orbits.
    • Engineering: Designing components like elliptical gears.
    • Architecture: Creating structures with elliptical shapes for aesthetic and functional purposes.

  12. 12.Eccentricity of an Ellipse

    Definition:

    Eccentricity (𝑒e) is a measure of how much an ellipse deviates from being a circle. It is a number between 0 and 1 for an ellipse. The closer the eccentricity is to 0, the more the ellipse resembles a circle. The closer it is to 1, the more elongated the ellipse becomes.

    Formula: For an ellipse, the eccentricity 𝑒e is given by the formula: 𝑒=𝑐𝑎e=ac​

    Where:

    • 𝑐c is the distance from the center to one of the foci.
    • 𝑎a is the length of the semi-major axis.

    Since 𝑐=𝑎2−𝑏2c=a2−b2​ (where 𝑏b is the semi-minor axis), the formula for eccentricity can also be written as: 𝑒=𝑎2−𝑏2𝑎e=aa2−b2​​

    Standard Equations of an Ellipse

    There are two standard forms of the equation of an ellipse, depending on whether the major axis is horizontal or vertical.

    1. Horizontal Major Axis

    The standard form of the equation of an ellipse with the center at the origin (0,0)(0,0) and the major axis along the x-axis is: 𝑥2𝑎2+𝑦2𝑏2=1a2x2​+b2y2​=1

    Where:

    • 𝑎a is the length of the semi-major axis.
    • 𝑏b is the length of the semi-minor axis.

    In this form:

    • The foci are at (±𝑐,0)(±c,0), where 𝑐=𝑎2−𝑏2c=a2−b2​.
    • The vertices are at (±𝑎,0)(±a,0).

    2. Vertical Major Axis

    The standard form of the equation of an ellipse with the center at the origin (0,0)(0,0) and the major axis along the y-axis is: 𝑥2𝑏2+𝑦2𝑎2=1b2x2​+a2y2​=1

    Where:

    • 𝑎a is the length of the semi-major axis.
    • 𝑏b is the length of the semi-minor axis.

    In this form:

    • The foci are at (0,±𝑐)(0,±c), where 𝑐=𝑎2−𝑏2c=a2−b2​.
    • The vertices are at (0,±𝑎)(0,±a).

    Example Calculation

    Let's consider an ellipse with 𝑎=5a=5 and 𝑏=3b=3.

    1. Calculate 𝑐c:

      𝑐=𝑎2−𝑏2=52−32=25−9=16=4c=a2−b2​=52−32​=25−9​=16​=4

    2. Calculate the eccentricity 𝑒e:

      𝑒=𝑐𝑎=45=0.8e=ac​=54​=0.8

    So, the eccentricity of this ellipse is 0.8, indicating it is fairly elongated.

    Real-Life Application

    Orbit of Planets: In astronomy, the orbits of planets around the sun are elliptical. The eccentricity of these orbits determines how circular or elongated the orbits are. For example, Earth's orbit has a low eccentricity (close to 0), making it nearly circular, while some comets have high eccentricities, resulting in highly elongated orbits.

    Careers and Industries

    Understanding eccentricity and the properties of ellipses is important in fields such as:

    • Astronomy: For studying the orbits of celestial bodies.
    • Engineering: For designing elliptical gears and arches.
    • Architecture: For creating aesthetically pleasing and structurally sound designs.

  13. 13.Latus rectum of an ellipse

    Definition and Key Elements

    Latus Rectum: The latus rectum of an ellipse is a line segment perpendicular to the major axis that passes through one of the foci and whose endpoints lie on the ellipse. Each focus has its own latus rectum, and these segments are equal in length.

    Elements in the Diagram

    • O: Center of the ellipse.
    • F1 and F2: Foci of the ellipse.
    • A and B: Vertices on the major axis.
    • C and D: Points where the latus rectum intersects the ellipse.
    • X and Y: Major and minor axes, respectively.

    Length of the Latus Rectum

    The length of the latus rectum (2𝑙2l) is given by the formula: 2𝑙=2𝑏2𝑎2l=a2b2​

    Where:

    • 𝑎a is the length of the semi-major axis.
    • 𝑏b is the length of the semi-minor axis.

    Derivation of the Length of the Latus Rectum

    To derive the length of the latus rectum, consider the standard equation of the ellipse centered at the origin (0,0)(0,0) with the major axis along the x-axis: 𝑥2𝑎2+𝑦2𝑏2=1a2x2​+b2y2​=1

    For a point (𝑥,𝑦)(x,y) on the latus rectum, the x-coordinate is the same as the x-coordinate of the focus (±𝑐±c), where 𝑐=𝑎2−𝑏2c=a2−b2​. Therefore, we substitute 𝑥=±𝑐x=±c into the ellipse equation to find the corresponding 𝑦y-coordinates.

    Calculation Steps

    1. Substitute 𝑥=±𝑐x=±c into the ellipse equation:

      𝑐2𝑎2+𝑦2𝑏2=1a2c2​+b2y2​=1

    2. Solve for 𝑦2y2:

      𝑐2𝑎2+𝑦2𝑏2=1a2c2​+b2y2​=1𝑦2𝑏2=1−𝑐2𝑎2b2y2​=1−a2c2​𝑦2𝑏2=𝑎2−𝑐2𝑎2b2y2​=a2a2−c2​

      Since 𝑐2=𝑎2−𝑏2c2=a2−b2:

      𝑦2𝑏2=𝑎2−(𝑎2−𝑏2)𝑎2b2y2​=a2a2−(a2−b2)​𝑦2𝑏2=𝑏2𝑎2b2y2​=a2b2​𝑦2=𝑏4𝑎2y2=a2b4​𝑦=±𝑏2𝑎y=±ab2​

    3. Length of the latus rectum: The total length of the latus rectum (from 𝐶C to 𝐷D) is:

      2𝑙=2×𝑏2𝑎=2𝑏2𝑎2l=2×ab2​=a2b2​

    Real-Life Application

    Optics: In optical systems, the concept of the latus rectum is used in the design of elliptical mirrors and lenses to focus light more effectively. The latus rectum helps in determining the focusing properties of these optical elements.

    Careers and Industries

    Understanding the latus rectum of an ellipse is useful in:

    • Astronomy: For studying the orbits of celestial bodies.
    • Engineering: In the design of elliptical gears and arches.
    • Optics: For designing elliptical mirrors and lenses

  14. 14.Exercise Questions

    Let's find the coordinates of the foci, the vertices, the lengths of the major and minor axes, the eccentricity, and the length of the latus rectum for each given ellipse.

    Problem 1: 𝑥236+𝑦216=136x2​+16y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=36a2=36 𝑎=36=6a=36​=6

    2. Semi-minor axis (b): 𝑏2=16b2=16 𝑏=16=4b=16​=4

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=36−16=20=25c=a2−b2​=36−16​=20​=25​

    4. Coordinates of the foci: (±25,0)(±25​,0)

    5. Vertices: (±6,0)(±6,0)

    6. Length of the major axis: 2𝑎=2×6=122a=2×6=12

    7. Length of the minor axis: 2𝑏=2×4=82b=2×4=8

    8. Eccentricity (e): 𝑒=𝑐𝑎=256=53e=ac​=625​​=35​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×166=326=163a2b2​=62×16​=632​=316​

    Problem 2: 𝑥24+𝑦225=14x2​+25y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=25a2=25 𝑎=25=5a=25​=5

    2. Semi-minor axis (b): 𝑏2=4b2=4 𝑏=4=2b=4​=2

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=25−4=21c=a2−b2​=25−4​=21​

    4. Coordinates of the foci: (0,±21)(0,±21​)

    5. Vertices: (0,±5)(0,±5)

    6. Length of the major axis: 2𝑎=2×5=102a=2×5=10

    7. Length of the minor axis: 2𝑏=2×2=42b=2×2=4

    8. Eccentricity (e): 𝑒=𝑐𝑎=215e=ac​=521​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×45=85a2b2​=52×4​=58​

    Problem 3: 𝑥216+𝑦29=116x2​+9y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=16a2=16 𝑎=16=4a=16​=4

    2. Semi-minor axis (b): 𝑏2=9b2=9 𝑏=9=3b=9​=3

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=16−9=7c=a2−b2​=16−9​=7​

    4. Coordinates of the foci: (±7,0)(±7​,0)

    5. Vertices: (±4,0)(±4,0)

    6. Length of the major axis: 2𝑎=2×4=82a=2×4=8

    7. Length of the minor axis: 2𝑏=2×3=62b=2×3=6

    8. Eccentricity (e): 𝑒=𝑐𝑎=74e=ac​=47​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×94=184=4.5a2b2​=42×9​=418​=4.5

    Problem 4: 𝑥225+𝑦2100=125x2​+100y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=100a2=100 𝑎=100=10a=100​=10

    2. Semi-minor axis (b): 𝑏2=25b2=25 𝑏=25=5b=25​=5

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=100−25=75=53c=a2−b2​=100−25​=75​=53​

    4. Coordinates of the foci: (0,±53)(0,±53​)

    5. Vertices: (0,±10)(0,±10)

    6. Length of the major axis: 2𝑎=2×10=202a=2×10=20

    7. Length of the minor axis: 2𝑏=2×5=102b=2×5=10

    8. Eccentricity (e): 𝑒=𝑐𝑎=5310=32e=ac​=1053​​=23​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×2510=5010=5a2b2​=102×25​=1050​=5

    Problem 5: 𝑥249+𝑦236=149x2​+36y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=49a2=49 𝑎=49=7a=49​=7

    2. Semi-minor axis (b): 𝑏2=36b2=36 𝑏=36=6b=36​=6

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=49−36=13c=a2−b2​=49−36​=13​

    4. Coordinates of the foci: (±13,0)(±13​,0)

    5. Vertices: (±7,0)(±7,0)

    6. Length of the major axis: 2𝑎=2×7=142a=2×7=14

    7. Length of the minor axis: 2𝑏=2×6=122b=2×6=12

    8. Eccentricity (e): 𝑒=𝑐𝑎=137e=ac​=713​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×367=727a2b2​=72×36​=772​

    Problem 6: 𝑥2100+𝑦2400=1100x2​+400y2​=1

    Solution:

    1. Semi-major axis (a): 𝑎2=400a2=400 𝑎=400=20a=400​=20

    2. Semi-minor axis (b): 𝑏2=100b2=100 𝑏=100=10b=100​=10

    3. Distance of the foci from the center (c): 𝑐=𝑎2−𝑏2=400−100=300=103c=a2−b2​=400−100​=300​=103​

    4. Coordinates of the foci: (0,±103)(0,±103​)

    5. Vertices: (0,±20)(0,±20)

    6. Length of the major axis: 2𝑎=2×20=402a=2×20=40

    7. Length of the minor axis: 2𝑏=2×10=202b=2×10=20

    8. Eccentricity (e): 𝑒=𝑐𝑎=10320=32e=ac​=20103​​=23​​

    9. Length of the latus rectum: 2𝑏2𝑎=2×10020=20020=10a2b2​=202×100​=20200​=10

  15. 15.Hyperbola

    Hyperbola Definition

    A hyperbola is a type of conic section that is formed when a plane intersects both nappes (the upper and lower parts) of a double cone. It consists of two separate curves, called branches, which are mirror images of each other and open in opposite directions.

    Eccentricity Definition

    Eccentricity (denoted as 𝑒e) is a measure of how much a conic section deviates from being circular. For a hyperbola, the eccentricity is always greater than 1. It is defined as the ratio of the distance from any point on the hyperbola to its focus and the perpendicular distance from that point to the nearest directrix.

    𝑒=𝑐𝑎e=ac​

    where:

    • 𝑐c is the distance from the center to the focus
    • 𝑎a is the distance from the center to a vertex

    Standard Equation of Hyperbola

    There are two standard forms of the equation of a hyperbola, depending on the orientation of the transverse axis (the axis that passes through the vertices of the hyperbola).

    1. Horizontal Transverse Axis: 𝑥2𝑎2−𝑦2𝑏2=1a2x2​−b2y2​=1

    2. Vertical Transverse Axis: 𝑦2𝑎2−𝑥2𝑏2=1a2y2​−b2x2​=1

    In both cases:

    • (ℎ,𝑘)(h,k) is the center of the hyperbola.
    • 𝑎a is the distance from the center to a vertex.
    • 𝑏b is related to 𝑎a and 𝑐c by the equation 𝑏2=𝑐2−𝑎2b2=c2−a2.
    • 𝑐c is the distance from the center to each focus.

    Latus Rectum Definition

    The latus rectum of a hyperbola is a line segment perpendicular to the transverse axis and passing through a focus. Its length is given by 2𝑏2𝑎a2b2​.

    Problem Solving Examples

    Example 1: Finding the Equation of a Hyperbola

    Problem: Find the equation of a hyperbola with foci at (±5, 0) and vertices at (±3, 0).

    Solution:

    1. Determine 𝑐c from the foci: 𝑐=5c=5.
    2. Determine 𝑎a from the vertices: 𝑎=3a=3.
    3. Calculate 𝑏b using 𝑏2=𝑐2−𝑎2b2=c2−a2: 𝑏2=52−32=25−9=16b2=52−32=25−9=16 𝑏=4b=4
    4. The equation is: 𝑥232−𝑦242=132x2​−42y2​=1 𝑥29−𝑦216=19x2​−16y2​=1

    Example 2: Eccentricity of a Hyperbola

    Problem: Calculate the eccentricity of a hyperbola given by the equation 𝑥29−𝑦216=19x2​−16y2​=1.

    Solution:

    1. Identify 𝑎a and 𝑏b: 𝑎2=9⇒𝑎=3a2=9⇒a=3 𝑏2=16⇒𝑏=4b2=16⇒b=4
    2. Calculate 𝑐c: 𝑐2=𝑎2+𝑏2=9+16=25c2=a2+b2=9+16=25 𝑐=5c=5
    3. Calculate eccentricity 𝑒e: 𝑒=𝑐𝑎=53e=ac​=35​

    Example 3: Length of the Latus Rectum

    Problem: Find the length of the latus rectum of the hyperbola 𝑥216−𝑦29=116x2​−9y2​=1.

    Solution:

    1. Identify 𝑎a and 𝑏b: 𝑎2=16⇒𝑎=4a2=16⇒a=4 𝑏2=9⇒𝑏=3b2=9⇒b=3
    2. Length of the latus rectum: 2𝑏2𝑎=2×94=184=4.5a2b2​=42×9​=418​=4.5

    Example 4: Vertices and Foci of a Hyperbola

    Problem: For the hyperbola 𝑦225−𝑥216=125y2​−16x2​=1, find the vertices and foci.

    Solution:

    1. Identify 𝑎a and 𝑏b: 𝑎2=25⇒𝑎=5a2=25⇒a=5 𝑏2=16⇒𝑏=4b2=16⇒b=4
    2. Vertices: Since the transverse axis is vertical, vertices are at (0, ±5).
    3. Calculate 𝑐c: 𝑐2=𝑎2+𝑏2=25+16=41c2=a2+b2=25+16=41 𝑐=41c=41​
    4. Foci: Foci are at (0, ±√41).

    Example 5: Asymptotes of a Hyperbola

    Problem: Find the equations of the asymptotes for the hyperbola 𝑦236−𝑥216=136y2​−16x2​=1.

    Solution:

    1. Identify 𝑎a and 𝑏b: 𝑎2=36⇒𝑎=6a2=36⇒a=6 𝑏2=16⇒𝑏=4b2=16⇒b=4
    2. Asymptotes: For vertical transverse axis, the equations are 𝑦=±𝑎𝑏𝑥y=±ba​x: 𝑦=±64𝑥y=±46​x 𝑦=±32𝑥y=±23​x

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