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The conic section is a curve which is formed by intersecting a plane with a cone. The curves that are formed are circle, ellipse, parabola and hyperbola. These curves are used in various fields for the designing of different types of equipment like antennas, telescopes, reflectors, etc.
Chapter 11 Conic Sections is one of the most important chapters of CBSE Class 11 Mathematics Syllabus. The Chapter falls under the Unit-3 i.e Coordinate Geometry of NCERT Class 11 Mathematics. The whole unit 3 consists of a total of 3 chapters and carries a combined weightage of 10 Marks in the CBSE Class 11 Mathematics Examination.
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Conic Sections
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When a plane intersects a cone in different sections, several types of curves are obtained. These curves can be circle, ellipse, parabola and hyperbola. When a plane cuts the cone other than the vertex following situations are expected:
Let β is the angle made by the plane with the vertical axis of the cone
(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
The four sections of a cone are explained below:
- Circle- It is the locus of a point which is moving in a certain plane around a fixed distance. The equation of a circle with centre (h, k) and the radius r is-
(x – h)2 + (y – k)2 = r2
- (x – h)2 + (y – k)2 = r2, is the equation of a circle with radius r having centre (h, k)
- x2 + y2 + 2gx + 2fy + c = 0 is the general equation of the circle, where, g, f and c are constants.
- x2 + y2 + 2gx + 2fy = 0, is the general equation of the circle passing through origin.
- 2g+2f-c, is the radius of the circle, r. The centre of the circle is (-g, -f).
- Ellipse- Ellipse is the set of all points in a plane, the sum of whose distance from two fixed points called its foci in a plane is constant. The equation of an ellipse with foci on the x-axis is
x²a²+ y²b² = 1
- Parabola- A parabola is a set of points in a plane whose distance from a fixed point is equal to the distance from the moving point to fixed straight lines. The equation of the parabola with focus at (a, 0) a > 0 and directrix x = – a is
y2 = 4ax
- Hyperbola- A hyperbola is the set of all points in a plane, the difference of whose distance is, from two fixed points in the plane, constant. The equation of a hyperbola with foci on the x-axis is :
x²a²- y²b² = 1
Discover about the Chapter video:
Conic Sections Detailed Video Explanation:
Read More: Conic Sections in Detail ( PDF to embed https://nios.ac.in/media/documents/SrSec311NEW/311_Maths_Eng/311_Maths_Eng_Lesson16.pdf)
Important Conic parameters
- Latus Rectum: It is a chord which is parallel to the directrix and passes through a focus.
- Directrix: A conic section is defined as the locus of a point P moving in the plane of a fixed point F known as focus (F) and a fixed line d known as directrix (with the focus not on d) such that the ratio of point P's distance from focus F to its distance from d equals a constant e known as eccentricity.
- Eccentricity: Eccentricity is a measure of how far the ellipse deviates from being circular. The eccentricity is if the angle formed between the cone's surface and its axis is and the angle formed between the cutting plane and the axis is: e = cos α/cos β.
- Axis: Any line which passes through the focus and is perpendicular to the directrix is the axis.
- Principal axis: The line connecting the two focal points or foci of an ellipse or hyperbola is known as the principal axis. The curve's centre is halfway.
- The major axis: It is a chord that connects the two vertices. It is an ellipse's longest chord.
- Minor axis: The ellipse's shortest chord is called the minor axis.
- Linear Eccentricity: It is the distance between a section's focus and its centre.
- Focal parameter: The distance between the focus and the accompanying directrix is known as the focal parameter.
Formulas:
Given below are the different types of conic sections and their formulas:
- Circle:
(x−a)2+(y−b)2=r2,
Where (a, b) is the center and r is the radius
- Horizontal major axis ellipse:
(x−a)2/h2+(y−b)2/k2=1
Where, (a, b) is the center, the length of the major axis is 2h, Length of the minor axis is 2k distance between the centre and either focus is c with c2=h2−k2, h>k>0
- Vertical major axis ellipse:
(x−a)2/k2+(y−b)2/h2=1
Where, (a, b) is the center, length of the major axis is 2h, length of the minor axis is 2k, distance between the centre and either focus is c with c2=h2−k2, h>k>0.
- Horizontal transverse axis hyperbola:
(x−a)2/h2−(y−b)2/k2=1
Where, (a, b) is the center, distance between the vertices is 2h, distance between the foci is 2k, c2= h2 + k2.
- Vertical transverse axis hyperbola:
(x−a)2/k2−(y−b)2/h2=1
Where, (a, b) is the center, distance between the vertices is 2h, distance between the foci is 2k, c2= h2 + k2
- Horizontal axis parabola:
(y−b)2=4p(x−a)
Where (a, b) is the vertex, focus is (a+p,b), directrix is the line, x=a−p, axis is the line y=b and p≠0
- Vertical axis parabola:
(x−a)2=4p(y−b)
Where (a, b) is the vertex, focus is (a+p,b), directrix is the line, x=b−p, axis is the line x=a and p≠0
Identifying a conic section:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is the standard form of equation of a conic section, where the real numbers are: A, B, C, D, E, F with A ≠ 0, B ≠ 0, C ≠ 0.
- If A = C and B = 0, then the conic section is a circle.
- If B2 – 4AC < 0, then the conic section is an ellipse.
- If B2 – 4AC = 0, then the conic section is a parabola
- If B2 – 4AC > 0, then the conic section is a hyperbola.
- If eccentricity, e = 0, the conic is a circle
- If 0, the conic is an ellipse
- If e=1, the conic is a parabola
- And if e>1, it is a hyperbola
Sample Questions Based on Conic Sections:
Ques.1: Find the equation of circle with center (-2, 3) and radius 4 (2 marks)
Answer: Given that: h=-2, k=3, r=4
The standard equation of Circle is-
(x-h)2+ (y-k)2= r2
By substituting the values
(x+2)2+ (y-3)2= 42
x2 + 4 + 4x+ y2+ 9- 6y= 16
x2 + y2+4x-6y-3=0
Ques.2: Find the center and radius of the circle with equation (x+5)2 + (y-3)2= 36 (2 marks)
Answer: The given equation is (x+5)2 + (y-3)2= 36
(x+5)2 + (y-3)2=62
(x-(-5))2 + (y-3)2=62
On comparing this with the standard equation of circle. We get,
h= -5, k=3 and r= 6
Hence the center lies at (-5,3) and the radius is 6.
Ques.3: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y2 = 12x. (2 marks)
Answer: The given equation is y2 = 12x. Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y2 = 4ax, we obtain 4a = 12 ⇒ a = 3
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y2, the axis of the parabola is the x-axis.
Equation of directrix, x = –a i.e., x = – 3
i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12.
Ques.4: Find the coordinates of the foci, the vertice, the length of major axis and minor axis, the eccentricity and the length of the latus rectum of the ellipse with the equation.(2 marks)
x²16+ y²9= 1.
Answer: Given that the equation of Ellipse-
x²16+ y²9= 1
x²4²+ y²3²= 1
On comparing with the standard equation. We get,
=> a=4 and b=3
Length of major axis = 2a= 8 units
Length of minor axis = 2b= 6 units
Eccentricity e = 1-b²a² = 1-916 = 7/ 4
Coordinates of foci = (±7 , 0)
Coordinates of vertices = (±4,0)
Latus Rectum = 2b²/a= 92
Ques.5: Find the equation of the hyperbola whose foci are (0, ±12) and the length of the latus rectum is 36.(2 marks)
Answer: Given foci are (0, ± 12)
That means c = 12.
Length of the latus rectum = 2b2/a = 36 or b2 = 18a
Now, c2 = a2 + b2
144 = a2 + 18a
i.e., a2 + 18a – 144 = 0
⇒ a = – 24, 6
The value of a cannot be negative.
Therefore, a = 6 and so b2 = 108.
Hence, the equation of the required hyperbola is (y2/36) – (x2/108) = 1 or 3y2 – x2 = 108.
Ques.6: Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.(3 marks)
Answer:
Let us consider the equation of the required circle be (x – h)2 + (y – k)2 = r2
We know that the circle passes through points (2,3) and (-1,1).
(2 – h)2+ (3 – k)2 =r2 ….Eq(1)
(-1 – h)2+ (1– k)2 =r2 ….Eq(2)
Since, the centre (h, k) of the circle lies on line x – 3y – 11= 0,
h – 3k =11….Eq(3)
From the equation (1) and (2), we obtain
(2 – h)2+ (3 – k)2 =(-1 – h)2 + (1 – k)2
4 – 4h + h2 +9 -6k +k2 = 1 + 2h +h2+1 – 2k + k2
4 – 4h +9 -6k = 1 + 2h + 1 -2k
6h + 4k =11….Eq(4)
Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,
6h+ 4k – 6(h-3k) = 11 – 66
6h + 4k – 6h + 18k = 11 – 66
22 k = – 55
K = -5/2
Substitute this value of K in equation (4) to get,
6h + 4(-5/2) = 11
6h – 10 = 11
6h = 21
h = 21/6
h = 7/2
We obtain h = 7/2and k = -5/2
On substituting the values of h and k in equation (1), we get
(2 – 7/2)2 + (3 + 5/2)2 = r2
[(4-7)/2]2 + [(6+5)/2]2 = r2
(-3/2)2 + (11/2)2 = r2
9/4 + 121/4 = r2
130/4 = r2
The equation of the required circle is
(x – 7/2)2 + (y + 5/2)2 = 130/4
[(2x-7)/2]2 + [(2y+5)/2]2 = 130/4
4x2 -28x + 49 +4y2 + 20y + 25 =130
4x2 +4y2 -28x + 20y – 56 = 0
4(x2 +y2 -7x + 5y – 14) = 0
x2 + y2 – 7x + 5y – 14 = 0
∴ The equation of the required circle is x2 + y2 – 7x + 5y – 14 = 0
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