Class 11 · Maths · Chapter 10 · NCERT Class 11 Mathematics

Conic Sections Class 11 Notes

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Chapter mind map

The whole chapter at a glance: the big idea, then each branch and what sits under it.

Conic Sections

Curves formed by the intersection of a plane with a double-napped right circular cone, varying by the intersection angle beta.

  1. The Circle

    Set of all points in a plane equidistant from a fixed centre point (h, k).

    • Standard Equation — Expressed as (x - h)² + (y - k)² = r², simplifying to x² + y² = r² when the centre is at the origin.
    • Radius and Centre — The fixed distance is radius 'r'; centre is found by completing the square in general quadratic forms.
  2. The Parabola

    Points equidistant from a fixed focus and a fixed line called the directrix.

    • Standard Forms — Equations y² = ±4ax (horizontal axis) or x² = ±4ay (vertical axis) based on orientation.
    • Latus Rectum — A segment perpendicular to the axis through the focus with length 4a.
  3. The Ellipse

    Points where the sum of distances from two fixed foci is a constant 2a.

    • Geometric Constants — Relationship a² = b² + c² where 2a is major axis, 2b is minor axis, and 2c is distance between foci.
    • Eccentricity and Latus Rectum — Eccentricity e = c/a is always less than 1; latus rectum length is 2b²/a.
  4. The Hyperbola

    Points where the difference of distances from two fixed foci is a constant 2a.

    • Transverse and Conjugate Axes — Transverse axis (2a) connects vertices; conjugate axis (2b) is perpendicular; c² = a² + b².
    • Eccentricity and Orientation — e = c/a > 1. Positive term in equation (x²/a² or y²/a²) determines the transverse axis direction.
  5. Conic Generation

    Curves defined by the angle beta of the plane relative to the cone's generator angle alpha.

    • Standard Conics — Circle (beta=90), Parabola (beta=alpha), Ellipse (alpha<beta<90), and Hyperbola (0<=beta<alpha).
    • Degenerate Cases — Occur when the plane passes through the vertex, resulting in a point, line, or intersecting lines.
  6. Real-world Applications

    Practical uses of conic properties in engineering, physics, and astronomy.

    • Reflective Properties — Parabolic reflectors in headlights and telescopes use the property that rays from the focus reflect parallel to the axis.
    • Orbits and Arches — Planetary orbits follow elliptical paths; elliptical and parabolic shapes are used in architectural arches.

Chapter notes

An exploration of curves formed by the intersection of a plane with a double-napped right circular cone, including circles, parabolas, ellipses, and hyperbolas, along with their standard equations and geometric properties.

Introduction to Conic Sections

Conic sections are curves obtained by the intersection of a plane with a double-napped right circular cone. These curves have significant applications in physics and engineering, such as planetary orbits and the design of reflectors.

A double-napped right circular cone is generated by rotating a line 'm' (the generator) around a fixed vertical axis 'l' at a constant angle α. The point of intersection is the vertex 'V'. The vertex divides the cone into two parts called nappes.

The type of curve produced depends on the angle β that the intersecting plane makes with the vertical axis. If the plane does not pass through the vertex, we obtain standard conics. If it passes through the vertex, we obtain 'degenerate' conics such as a point, a single line, or a pair of intersecting lines.

The four primary conic sections are the circle (β = 90°), the ellipse (α < β < 90°), the parabola (β = α), and the hyperbola (0 ≤ β < α). In the case of the hyperbola, the plane cuts through both nappes of the cone.

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Question

What determines the specific type of conic section formed by a plane and a cone?

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NCERT reference: chapter PDF pages 1, 2, 3.

The Circle

A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the centre.

The fixed distance from the centre to any point on the circle is known as the radius 'r'. If the centre of the circle is at the point C(h, k) and P(x, y) is any point on the circle, the distance CP must equal r.

Using the distance formula, we have √((x - h)² + (y - k)²) = r. Squaring both sides gives the standard equation: (x - h)² + (y - k)² = r². If the centre is at the origin (0, 0), the equation simplifies to x² + y² = r².

Finding the Equation of a Circle

(x + 3)² + (y - 2)² = 16

For a circle with centre (-3, 2) and radius 4, we substitute h = -3, k = 2, and r = 4 into the standard form (x - h)² + (y - k)² = r². This gives (x - (-3))² + (y - 2)² = 4², which simplifies to (x + 3)² + (y - 2)² = 16.

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Question

What is the centre and radius of the circle given by x² + y² + 8x + 10y - 8 = 0?

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NCERT reference: chapter PDF pages 4, 5.

The Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed line (the directrix) and a fixed point (the focus) not on the line.

The line through the focus and perpendicular to the directrix is the axis of the parabola. The point where the parabola intersects its axis is the vertex. The distance from the vertex to the focus is denoted as 'a'.

Standard equations depend on the orientation. For a parabola opening to the right with vertex at (0,0) and focus at (a, 0), the equation is y² = 4ax. If it opens to the left, y² = -4ax. If it opens upwards, x² = 4ay, and if downwards, x² = -4ay.

EquationFocusDirectrixAxis
y² = 4ax(a, 0)x = -ax-axis
y² = -4ax(-a, 0)x = ax-axis
x² = 4ay(0, a)y = -ay-axis
x² = -4ay(0, -a)y = ay-axis

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Question

Define the latus rectum of a parabola.

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NCERT reference: chapter PDF pages 7, 8, 9.

Worked Example: Parabola Properties

Calculating the properties of a parabola from its standard equation.

To find the focus and directrix, we compare the given equation to the standard form. The coefficient of the linear term represents 4a.

Analyzing y² = 8x

y² = 8x

Comparing y² = 8x with y² = 4ax, we get 4a = 8, so a = 2. The axis of symmetry is the x-axis because the equation involves y². Since the coefficient is positive, it opens right. Focus is (a, 0) = (2, 0). Directrix is x = -a, so x = -2. Length of latus rectum is 4a = 8.

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Question

Find the equation of a parabola with vertex (0,0) and focus (0, 2).

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NCERT reference: chapter PDF pages 10, 11.

The Ellipse

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (foci) is a constant.

The line segment through the foci is the major axis (length 2a), and the segment perpendicular to it through the centre is the minor axis (length 2b). The distance between the foci is 2c.

A fundamental relationship exists between these constants: a² = b² + c², which can be expressed as c = √(a² - b²). The eccentricity 'e' is the ratio c/a. Since c < a for an ellipse, the eccentricity is always less than 1.

Ellipse Components

  1. 1

    Foci

    Two fixed points F1 and F2 located on the major axis.

  2. 2

    Major Axis

    The longest diameter, passing through the foci and centre.

  3. 3

    Minor Axis

    The shortest diameter, perpendicular to the major axis at the centre.

The geometric structure of an ellipse defined by its axes and foci.

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What is the constant sum of distances for any point on an ellipse?

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NCERT reference: chapter PDF pages 12, 13.

Standard Equations of an Ellipse

The equation of an ellipse is simplest when the centre is at the origin and the axes lie along the coordinate axes.

If the major axis is along the x-axis, the equation is (x²/a²) + (y²/b²) = 1, where a > b. If the major axis is along the y-axis, the equation is (x²/b²) + (y²/a²) = 1, where a > b.

To identify the major axis from an equation, look at the denominators. The larger denominator is always a², and the axis corresponding to that variable is the major axis.

Analyzing an Ellipse Equation

(x²/25) + (y²/9) = 1

Here a² = 25 and b² = 9, so a = 5 and b = 3. Since the larger denominator is under x², the major axis is along the x-axis. c² = a² - b² = 25 - 9 = 16, so c = 4. Foci are (±4, 0), vertices are (±5, 0). Eccentricity e = c/a = 4/5. Latus rectum length = 2b²/a = 2(9)/5 = 18/5.

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Question

Find the equation of the ellipse with vertices (0, ±13) and foci (0, ±5).

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NCERT reference: chapter PDF pages 14, 15, 16.

The Hyperbola

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points (foci) is a constant.

The line through the foci is the transverse axis (length 2a), and the line through the centre perpendicular to it is the conjugate axis (length 2b). The distance between foci is 2c.

For a hyperbola, c² = a² + b². The eccentricity e = c/a is always greater than 1 because c > a. The points where the hyperbola intersects the transverse axis are the vertices.

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How does the definition of a hyperbola differ from an ellipse?

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NCERT reference: chapter PDF pages 20, 21, 22, 23, 24.

Standard Equations of a Hyperbola

Standard forms of hyperbola equations depend on whether the transverse axis is horizontal or vertical.

If the transverse axis is along the x-axis, the equation is (x²/a²) - (y²/b²) = 1. If the transverse axis is along the y-axis, the equation is (y²/a²) - (x²/b²) = 1.

The positive term in the equation indicates the direction of the transverse axis. The denominator of the positive term is always a², which defines the length of the semi-transverse axis.

Analyzing a Hyperbola

(x²/9) - (y²/16) = 1

Comparing with (x²/a²) - (y²/b²) = 1, we have a² = 9 and b² = 16, so a = 3 and b = 4. c = √(a² + b²) = √(9 + 16) = 5. Foci are (±5, 0), vertices are (±3, 0). Eccentricity e = c/a = 5/3. Latus rectum = 2b²/a = 2(16)/3 = 32/3.

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What is an equilateral hyperbola?

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NCERT reference: chapter PDF pages 23, 24, 25.

Applications and Miscellaneous Examples

Conic sections are not just theoretical; they describe physical phenomena and are used in engineering designs.

Parabolic reflectors are used in headlights and telescopes because light rays originating from the focus are reflected parallel to the axis. Elliptical shapes are found in arches and the orbits of planets.

When solving real-world problems, it is crucial to set up a coordinate system where the vertex or centre is at the origin to simplify the equations.

Parabolic Mirror Depth

y² = 20x

A parabolic mirror has its focus 5 cm from the vertex. Thus a = 5, and the equation is y² = 4(5)x = 20x. If the mirror is 45 cm deep, we set x = 45. Then y² = 20(45) = 900, so y = ±30. The diameter AB is 2y = 60 cm.

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A rod of length 15 cm rests between coordinate axes. If a point P on the rod is 6 cm from the end on the x-axis, what is the locus of P?

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NCERT reference: chapter PDF pages 27, 28, 29.

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