Conic Sections Class 11
In conic Sections Class 11, we will study about different kinds of curves like circles, ellipse, hyperbola and parabolas. The curves are known as conic sections or conics. Because the curves are obtained from the intersection of a plane with a double-napped right circular cone. These curves have a wide range of applications in various fields like automobile headlights, designing of antennas and telescope, reflectors etc. In conic sections, class 11 chapter 11 concepts help to understand the different sections of a cone when it is rotated.
Conic Sections Class 11 Concepts
The topics and subtopics covered in conic sections class 11 are:
- Introduction
- Sections of a Cone
- Circle, ellipse, parabola and hyperbola
- Degenerated conic sections
- Circle
- Parabola
- Standard equations of parabola
- Latus rectum
- Ellipse
- Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse
- Special cases of an ellipse
- Eccentricity
- Standard equations of an ellipse
- Latus rectum
- Hyperbola
- Eccentricity
- Standard equation of Hyperbola
- Latus rectum
Conic Sections Class 11 Notes
When a plane cuts the cone other than the vertex, we have the following situations:
(a) When β = 90°, the section is a circle
(b) When α < β < 90°, the section is an ellipse
(c) When α = β; the section is a parabola
(d) When 0 ≤ β < α; the section is a hyperbola
Where β is the angle made by the plane with the vertical axis of the cone.
Circle: Set of points in a plane equidistant from a fixed point. A circle with radius r and centre (h, k) can be represented as (x – h)^{2 }+ (y – k)^{2 }= r^{2}
Parabola: Set of points in a plane that are equidistant from a fixed-line and point. A parabola with a > 0, focus at (a, 0), and directrix x = – a can be represented as y^{2 }= 4ax
In parabola y^{2 }= 4ax, the length of the latus rectum is given by 4a.
Ellipse: The sum of distances of a set of points in a plane from two fixed points is constant. An ellipse with foci on the x-axis can be represented as: \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
In ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), the length of the latus rectum is given by \(\frac{2b^{2}}{a}\).
Hyperbola: The difference of distances of set of points in a plane from two fixed points is constant. The hyperbola with foci on the x-axis can be represented as:
\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)In a Hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), the length of latus rectum is given by \(\frac{2b^{2}}{a}\)
Conic Sections Class 11 Examples
Example 1: Find an equation of the circle with centre at (0, 0) and radius r.
Solution: Given,
Centre = (h, k) = (0, 0)
Radius = r
Therefore, the equation of the circle is x^{2} + y^{2} = r^{2}.
Example 2: Find the equation of the circle with centre (–3, 2) and radius 4.
Solution: Given,
Centre = (h, k) = (-3, 2)
Radius = r = 4
Therefore, the equation of the required circle is
(x – h)^{2} + (y – k)^{2} = r^{2}
(x + 3)^{2} + (y – 2)^{2} = 4^{2}
(x + 3)^{2} + (y – 2)^{2} = 16
Example 3: Find the centre and the radius of the circle x^{2} + y^{2} + 8x + 10y – 8 = 0
Solution: The given equation is
x^{2} + y^{2} + 8x + 10y – 8 = 0
(x^{2} + 8x) + (y^{2} + 10y) = 8
By completing the squares within the parenthesis, we get
(x^{2} + 8x + 16) + (y^{2} + 10y + 25) = 8 + 16 + 25
i.e. (x + 4)^{2} + (y + 5)2 = 49
i.e. [x – (– 4)]^{2} + [y – (–5)]^{2} = 7^{2}
Comparing with the standard form, h = -4, k = -5 and r = 7
Therefore, the given circle has centre at (– 4, –5) and radius 7.
Example 4: Find the equation of the circle which passes through the points (2, – 2), and (3, 4) and whose centre lies on the line x + y = 2.
Solution: Let the equation of the circle be (x – h)^{2} + (y – k)^{2} = r^{2}.
Given that the circle passes through the points (2, –2) and (3, 4).
Thus,
(2 – h)^{2} + (–2 – k)^{2} = r^{2}….(1)
and (3 – h)^{2} + (4 – k)^{2} = r^{2}….(2)
Also, given that the centre lies on the line x + y = 2.
⇒ h + k = 2 ….(3)
Solving the equations (1), (2) and (3), we get
h = 0.7, k = 1.3 and r^{2} = 12.58
Hence, the equation of the required circle is
(x – 0.7)^{2} + (y – 1.3)^{2} = 12.58
Example 5: Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola y^{2} = 8x.
Solution: The given equation involves y^{2}, that means the axis of symmetry is along the x-axis.
The coefficient of x is positive so the parabola opens to the right.
Now by comparing the given equation with y^{2} = 4ax,
a = 2
Thus, the focus of the parabola is (2, 0) and the equation of the directrix of the parabola is x = –2 (see the figure)
Length of the latus rectum = 4a = 4 × 2 = 8
Example 6: Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).
Solution: Given,
Vertex = (0,0)
Focus = (0,2)
The focus lies on the y-axis.
Thus, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is of the form x^{2} = 4ay.
⇒ x^{2} = 4(2)y
⇒ x^{2} = 8y
This is the required equation of parabola.
Example 7: Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse 9x^{2} + 4y^{2} = 36.
Solution: Given, 9x^{2} + 4y^{2} = 36
The given equation of the ellipse can be written in standard form as:
(x^{2}/4) + (y^{2}/9) = 1
Here, 9 is greater than 4.
Thus, the major axis is along the y-axis.
By comparing with (x^{2}/b^{2}) + (y^{2}/a^{2}) = 1,
a^{2} = 9, b^{2} = 4
⇒ a = 3, b = 2
c = √(a^{2} – b^{2}) = √(9 – 4) = √5
e = c/a = √5/3
Therefore, the foci are (0, √5) and (0, –√5), vertices are (0, 3) and (0, –3), length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the ellipse is √5/3.
Example 8: Find the equation of the hyperbola whose foci are (0, ±12) and the length of the latus rectum is 36.
Solution: Given foci are (0, ± 12)
That means c = 12.
Length of the latus rectum = 2b^{2}/a = 36 or b^{2} = 18a
Now, c^{2} = a^{2} + b^{2}
144 = a^{2} + 18a
i.e., a^{2} + 18a – 144 = 0
⇒ a = – 24, 6
The value of a cannot be negative.
Therefore, a = 6 and so b^{2} = 108.
Hence, the equation of the required hyperbola is (y^{2}/36) – (x^{2}/108) = 1 or 3y^{2} – x^{2} = 108.
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Also, read more class 11 concepts: | |
Binomial Theorem Class 11 | Complex Numbers Class 11 |
Limits and Derivatives Class 11 | Linear Inequalities Class 11 |