Conic Sections

NCERT Solutions for Class 11 Maths Chapter 10

Chapter 10.1: Introduction

• Conic Section: The curve formed by the intersection of a plane with a double cone.
• Types of Conic Sections: Circle, ellipse, parabola, and hyperbola.

Chapter 10.2: The Circle

• Equation of a Circle: (x – h)^2 + (y – k)^2 = r^2, where (h, k) is the center and r is the radius.
• Standard Form: x^2 + y^2 + 2gx + 2fy + c = 0
• Center and Radius: Center is (-g, -f) and radius is √(g^2 + f^2 – c).

Chapter 10.3: The Parabola

• Equation of a Parabola: y^2 = 4ax (vertical axis) or x^2 = 4ay (horizontal axis)
• Vertex: (0, 0)
• Axis: Line y = 0 (vertical axis) or x = 0 (horizontal axis)
• Focus: (a, 0) or (0, a)
• Directrix: x = -a or y = -a

Chapter 10.4: The Ellipse

• Equation of an Ellipse: x^2/a^2 + y^2/b^2 = 1 (standard form)
• Center: (0, 0)
• Vertices: (±a, 0) and (0, ±b)
• Foci: (±√(a^2 – b^2), 0) or (0, ±√(a^2 – b^2))
• Eccentricity: e = √(a^2 – b^2) / a

Chapter 10.5: The Hyperbola

• Equation of a Hyperbola: x^2/a^2 – y^2/b^2 = 1 (horizontal axis) or y^2/b^2 – x^2/a^2 = 1 (vertical axis)
• Center: (0, 0)
• Vertices: (±a, 0) or (0, ±b)
• Foci: (±√(a^2 + b^2), 0) or (0, ±√(a^2 + b^2))
• Eccentricity: e = √(a^2 + b^2) / a
• Asymptotes: y = ±(b/a)x

Key Concepts:

• Properties of conic sections
• Equations of conic sections
• Graphs of conic sections
• Applications of conic sections (e.g., optics, astronomy)

NCERT Solutions for Class 11 Maths Chapter 10

Exercise 10.1

In each of the following Exercises 1 to 5, find the equation of the circle with

1. centre (0,2) and radius 2

Ans:  

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (0, 2) and the radius is 2. Substituting these values into the equation:  

(x – 0)^2 + (y – 2)^2 = 2^2

Simplifying:

x^2 + y^2 – 4y + 4 = 4

x^2 + y^2 – 4y = 0

2. centre (–2,3) and radius 4

Ans : 

(x – h)^2 + (y – k)^2 = r^2

Substituting these values into the equation:  

(x – (-2))^2 + (y – 3)^2 = 4^2

(x + 2)^2 + (y – 3)^2 = 16

Expanding the equation:

x^2 + 4x + 4 + y^2 – 6y + 9 = 16

x^2 + y^2 + 4x – 6y – 3 = 0

3. Centre (1/2, 1/4 ) and radius 1/12

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (1/2, 1/4) and the radius is 1/12. Substituting these values into the equation:  

(x – 1/2)^2 + (y – 1/4)^2 = (1/12)^2

Expanding the equation:

x^2 – x + 1/4 + y^2 – y/2 + 1/16 = 1/144

Multiplying both sides by 144:

144x^2 – 144x + 36 + 144y^2 – 72y + 9 = 1

144x^2 + 144y^2 – 144x – 72y + 44 = 0

4. Centre (1,1) and radius 2

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (1, 1) and the radius is 2. Substituting these values into the equation:  

(x – 1)^2 + (y – 1)^2 = 2^2

Expanding the equation:

x^2 – 2x + 1 + y^2 – 2y + 1 = 4

x^2 + y^2 – 2x – 2y – 2 = 0

5. centre (–a, –b) and radius a2–b2

Ans : 

In each of the following Exercises 6 to 9, find the centre and radius of the circles.

6. (x + 5)2 + (y – 3)2 = 36

Ans : 

The given equation is already in the standard form of a circle:

(x – h)^2 + (y – k)^2 = r^2

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

Comparing the given equation with the standard form, we can see that:  

• h = -5
• k = 3
• r^2 = 36

Therefore, the center of the circle is (-5, 3) and its radius is √36 = 6 units.

7. x 2 + y 2 – 4x – 8y – 45 = 0

Ans : 

The given equation is:

x^2 + y^2 – 4x – 8y – 45 = 0

(x – h)^2 + (y – k)^2 = r^2

To do this, we can complete the square for both the x and y terms:  

x^2 – 4x + 4 + y^2 – 8y + 16 – 45 = 0 + 4 + 16

(x – 2)^2 + (y – 4)^2 – 25 = 0

(x – 2)^2 + (y – 4)^2 = 25

Comparing this equation to the standard form, we can see that:

• h = 2
• k = 4
• r^2 = 25

Therefore, the center of the circle is (2, 4) and its radius is √25 = 5 units.

8. x 2 + y 2 – 8x + 10y – 12 = 0

Ans : 

The given equation is:

x^2 + y^2 – 8x + 10y – 12 = 0

(x – h)^2 + (y – k)^2 = r^2

To do this, we can complete the square for both the x and y terms:  

x^2 – 8x + 16 + y^2 + 10y + 25 – 12 = 0 + 16 + 25

(x – 4)^2 + (y + 5)^2 – 43 = 0

(x – 4)^2 + (y + 5)^2 = 43

• h = 4
• k = -5
• r^2 = 43

Therefore, the center of the circle is (4, -5) and its radius is √43 units.

9. 2x 2 + 2y 2 – x = 0

Ans : 

To find the center and radius of the circle represented by the equation 2x^2 + 2y^2 – x = 0, we need to convert it into the standard form of a circle’s equation:

(x – h)^2 + (y – k)^2 = r^2

Steps:

1. Divide the entire equation by 2 to simplify:
   • x^2 + y^2 – x/2 = 0
2. Rearrange to group the x terms:
   • x^2 – x/2 + y^2 = 0
3. Complete the square for the x terms:
   • To complete the square, add and subtract (1/4)^2 (the square of half the coefficient of x):
     • x^2 – x/2 + (1/4)^2 – (1/4)^2 + y^2 = 0
4. Rewrite the equation:
   • (x – 1/4)^2 + y^2 = (1/4)^2

Now, the equation is in standard form.

Center: (1/4, 0) Radius: √(1/4)^2 = 1/4

Therefore, the center of the circle is (1/4, 0) and its radius is 1/4.

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

Ans :

NCERT Solutions for Class 11 Maths Chapter 10

11. 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.

Ans : 

Since the circle passes through the points (2, 3) and (-1, 1), we have:

(2 – h)^2 + (3 – k)^2 = (h + 1)^2 + (k – 1)^2

Expanding and simplifying:

h^2 – 4h + k^2 – 6k + 13 = h^2 + 2h + k^2 – 2k + 2

-6h – 4k + 11 = 0

3h + 2k – 11/2 = 0

Since the center lies on the line x – 3y – 11 = 0, we also have:

h – 3k – 11 = 0

Now, we have two equations:

3h + 2k – 11/2 = 0 h – 3k – 11 = 0

Solving these equations simultaneously, we get:

h = 7/2 k = -5/2

Therefore, the center of the circle is (7/2, -5/2).

To find the radius, we can use the distance between the center and any of the given points:

Radius = √((2 – 7/2)^2 + (3 – (-5/2))^2)

Radius = √((-1/2)^2 + (11/2)^2)

Radius = √(122/4)

Radius = √30.5

Finally, the equation of the circle is:

(x – 7/2)^2 + (y + 5/2)^2 = 30.5

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3). 

Ans : 

Let the center of the circle be (a, 0), since it lies on the x-axis.

The radius of the circle is given as 5 units.

Using the distance formula, we can write the equation:

(x – a)^2 + (y – 0)^2 = 5^2

(x – a)^2 + y^2 = 25

Since the circle passes through the point (2, 3), substituting these values in the equation:

(2 – a)^2 + 3^2 = 25

4 – 4a + a^2 + 9 = 25

a^2 – 4a – 12 = 0

Factoring the equation:

(a – 6)(a + 2) = 0

Therefore, the possible values for a are 6 and -2.

So, there are two circles that satisfy the given conditions:

1. Center: (6, 0), Radius: 5 Equation: (x – 6)^2 + y^2 = 25
2. Center: (-2, 0), Radius: 5 Equation: (x + 2)^2 + y^2 = 25

13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Ans : 

(0 – h)^2 + (0 – k)^2 = r^2

h^2 + k^2 = r^2

This means that the points (a, 0) and (0, b) lie on the circle.

Substituting (a, 0) into the equation of the circle:

(a – h)^2 + (0 – k)^2 = r^2

a^2 – 2ah + h^2 + k^2 = r^2

Since h^2 + k^2 = r^2 (from the first equation), we can substitute r^2:

a^2 – 2ah + r^2 = r^2

a^2 – 2ah = 0

a(a – 2h) = 0

Therefore, either a = 0 or a = 2h.

Similarly, substituting (0, b) into the equation of the circle:

(0 – h)^2 + (b – k)^2 = r^2

h^2 + b^2 – 2bk + k^2 = r^2

Since h^2 + k^2 = r^2, we can substitute r^2:

h^2 + b^2 – 2bk + r^2 = r^2

b^2 – 2bk = 0

b(b – 2k) = 0

Therefore, either b = 0 or b = 2k.

Now, we have four possible cases:

1. a = 0 and b = 0: This means the circle is centered at the origin, and its equation is x^2 + y^2 = r^2. Since it passes through (a, 0), the radius is a. So the equation is x^2 + y^2 = a^2.
2. a = 0 and b = 2k: In this case, the center is (0, k) and the radius is k. The equation of the circle is x^2 + (y – k)^2 = k^2.
3. a = 2h and b = 0: In this case, the center is (h, 0) and the radius is h. The equation of the circle is (x – h)^2 + y^2 = h^2.
4. a = 2h and b = 2k: In this case, the center is (h, k) and the radius is √(h^2 + k^2). The equation of the circle is (x – h)^2 + (y – k)^2 = h^2 + k^2.

Since the circle passes through the origin, the radius must be equal to the distance between the origin and the center. Therefore, in all cases, r = √(h^2 + k^2).

Combining the four cases, we can say that the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:

x^2 + y^2 – ax – by = 0

14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Ans : 

(x – h)^2 + (y – k)^2 = r^2

In this case, the center is (2, 2). To find the radius, we can use the distance formula between the center and the point (4, 5):  

r = √((4 – 2)^2 + (5 – 2)^2) = √(4 + 9) = √13

(x – 2)^2 + (y – 2)^2 = (√13)^2

(x – 2)^2 + (y – 2)^2 = 13

Expanding the equation:

x^2 – 4x + 4 + y^2 – 4y + 4 = 13

x^2 + y^2 – 4x – 4y – 5 = 0

Therefore, the equation of the circle passing through the point (4, 5) with center (2, 2) is:

x^2 + y^2 – 4x – 4y – 5 = 0

15. Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25? 

Ans : 

First, let’s identify the center and radius of the circle.

The equation x^2 + y^2 = 25 is in the standard form of a circle: (x – h)^2 + (y – k)^2 = r^2, where (h, k) is the center and r is the radius.

Comparing the given equation to the standard form, we can see that:  

• h = 0
• k = 0
• r = √25 = 5

So, the circle has its center at (0, 0) and a radius of 5.

Now, let’s determine the distance between the point (-2.5, 3.5) and the center (0, 0).

The distance formula is:

d = √((x2 – x1)^2 + (y2 – y1)^2)

Substituting the coordinates:

d = √((-2.5 – 0)^2 + (3.5 – 0)^2)

d = √(6.25 + 12.25)

d = √18.5

Since 18.5 is less than 25 (the square of the radius), the point (-2.5, 3.5) lies inside the circle.

Exercise 10.2

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

1. y2= 12x

Ans : 

General Form of a Parabola:

• y^2 = 4ax

• 4a = 12
• a = 3

Therefore, the parabola opens to the right.

Focus:

• The focus of a right-opening parabola is (a, 0).
• So, the focus is (3, 0).

Axis of the Parabola:

• Since the vertex is (0, 0), the axis of symmetry is the y-axis.
• Equation of the axis: x = 0

Directrix:

• The directrix of a right-opening parabola is a vertical line x = -a.
• So, the equation of the directrix is x 
• = -3.

Length of the Latus Rectum:

• So, the length of the latus rectum is 4 * 3 = 12.

Summary:

• Focus: (3, 0)
• Axis of the Parabola: x = 0
• Equation of the Directrix: x = -3
• Length of the Latus Rectum: 12

2. x2= 6y

Ans : 

3. y2= -8x

Ans :

Key points:

• Axis of symmetry: x-axis
• Vertex: (0, 0)
• Opens left (since the coefficient of x is negative)

To find the focus and directrix:

• 4a = -8 (comparing with the standard form y^2 = 4ax)
• a = -2
• Focus: (-a, 0) = (-2, 0)
• Directrix: x = a = 2

Therefore, the parabola y^2 = -8x has:

• Vertex: (0, 0)
• Axis of symmetry: x-axis
• Focus: (-2, 0)
• Directrix: x = 2

4. x2= -16y

Ans : 

5.  y2= 10x

Ans : 

The given equation is y^2 = 10x. This represents a parabola.

Key points:

• Axis of symmetry: y-axis
• Vertex: (0, 0)
• Opens right (since the coefficient of x is positive)

To find the focus and directrix:

• 4a = 10 (comparing with the standard form y^2 = 4ax)
• a = 5/2
• Focus: (a, 0) = (5/2, 0)
• Directrix: x = -a = -5/2

Therefore, the parabola y^2 = 10x has:

• Vertex: (0, 0)
• Axis of symmetry: y-axis
• Focus: (5/2, 0)
• Directrix: x = -5/2

6. x2= -9y

Ans : 

Key points:

• Axis of symmetry: y-axis
• Vertex: (0, 0)
• Opens downward (since the coefficient of y is negative)

To find the focus and directrix:

• 4a = -9 
• a = -9/4
• Focus: (0, -a) = (0, 9/4)
• Directrix: y = a = -9/4

Therefore, the parabola x^2 = -9y has:

• Vertex: (0, 0)
• Axis of symmetry: y-axis
• Focus: (0, 9/4)
• Directrix: y = -9/4

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus (6,0); directrix x = – 6

Ans :

Finding the Equation of the Parabola

Given:

• Focus: (6, 0)
• Directrix: x = -6

Analysis:

y^2 = 4ax

Finding the value of a:

• Vertex: ((6 + (-6))/2, 0) = (0, 0)
  
• This distance is the value of a.  
• a = 6
  

Substituting the value of a into the general equation:

y^2 = 4 * 6 * x

Therefore, the equation of the parabola is:

y^2 = 24x

8. Focus (0,–3); directrix y = 3

Ans : 

Given:

• Focus: (0, -3)
• Directrix: y = 3

Analysis:

x^2 = -4ay

Finding the value of a:

• Vertex: ((0 + 0)/2, (-3 + 3)/2) = (0, 0)
  
• This distance is the value of a.  
• a = 3
  

Substituting the value of a into the general equation:

x^2 = -4 * 3 * y

Therefore, the equation of the parabola is:

x^2 = -12y

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

Ans : 

Given:

• Vertex: (0, 0)
• Focus: (3, 0)

Analysis:

y^2 = 4ax

Finding the value of a:

• a = 3

Substituting the value of a into the general equation:

y^2 = 4 * 3 * x

Therefore, the equation of the parabola is:

y^2 = 12x

10.Vertex (0,0); focus (–2,0)

Ans : 

Given:

• Vertex: (0, 0)
• Focus: (-2, 0)

Analysis:

y^2 = -4ax

Finding the value of a:

• a = 2

Substituting the value of a into the general equation:

y^2 = -4 * 2 * x

Therefore, the equation of the parabola is:

y^2 = -8x

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

Ans : 

Given:

• Vertex: (0, 0)
• Axis: Along the x-axis
• Passes through the point (2, 3)

Analysis:

• Since the axis is along the x-axis, the parabola is of the form y^2 = 4ax.
• We need to find the value of a.

Substituting the given point (2, 3) into the equation:

3^2 = 4a * 2

9 = 8a

Solving for a:

a = 9/8

Therefore, the equation of the parabola is:

y^2 = 4 * (9/8) * x

Simplifying:

y^2 = 9/2 * x

NCERT Solutions for Class 11 Maths Chapter 10

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

Ans : 

Given:

• Vertex: (0, 0)
• Passes through the point (5, 2)
• Symmetric with respect to the y-axis

Analysis:

• Since the parabola is symmetric with respect to the y-axis and has its vertex at (0, 0), the general equation of the parabola is of the form x^2 = 4ay.

Substituting the given point (5, 2) into the equation:

5^2 = 4a * 2

25 = 8a

Solving for a:

a = 25 / 8

Substituting the value of a back into the general equation:

x^2 = 4 * (25/8) * y

Simplifying:

x^2 = 25/2 * y

Therefore, the equation of the parabola is:

x^2 = 25/2 * y

Exercise 10.3

1. In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse. 

1.

Ans : 

Analyzing the Ellipse: x^2/36 + y^2/16 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 36
• b^2 = 16

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √36 = 6
• Vertices: (±6, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(36 – 16) = √20 = 2√5
• The foci are (±c, 0).
• Foci: (±2√5, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a = 2 * 6 = 12

Finding the Length of the Minor Axis:

• Length of minor axis = 2b
•  = 2 * 4 
• = 8

Finding the Eccentricity:

• Eccentricity (e) = c / a = (2√5) / 6 = √5 / 3

Finding the Length of the Latus Rectum:

• Length of latus rectum = 2b^2 / a = 2 * 16 / 6 

= 16/3

2.

2. 

Ans : 

Analyzing the Ellipse: x^2/4 + y^2/25 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 4
• b^2 = 25

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
• The vertices are (0, ±b).
• b = √25 = 5
• Vertices: (0, ±5)

Finding the Foci:

• c = √(b^2 – a^2) = √(25 – 4) = √21
• The foci are (0, ±c).
• Foci: (0, ±√21)

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 5 = 10

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 2 = 4

Finding the Eccentricity:

• Eccentricity (e) = c / b = (√21) / 5

Finding the Length of the Latus Rectum:

Length of latus rectum = 2a^2 / b = 2 * 4 / 5 = 8/5

3.

Ans : 

Analyzing the Ellipse: x^2/16 + y^2/9 = 1

Standard Form of an Ellipse:

The given equation is already in standard form:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 16
• b^2 = 9

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √16 = 4
• Vertices: (±4, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(16 – 9) = √7
• The foci are (±c, 0).
• Foci: (±√7, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a = 2 * 4 = 8

Finding the Length of the Minor Axis:

• Length of minor axis = 2b = 2 * 3 = 6

Finding the Eccentricity:

• Eccentricity (e) = c / a = (√7) / 4

Finding the Length of the Latus Rectum:

Length of latus rectum = 2b^2 / a = 2 * 9 / 4 = 9/2

4.

Ans : 

Analyzing the Ellipse: x^2/25 + y^2/100 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 25
• b^2 = 100

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
• The vertices are (0, ±b).
• b = √100 = 10
• Vertices: (0, ±10)

Finding the Foci:

• c = √(b^2 – a^2) = √(100 – 25) = √75 = 5√3
• The foci are (0, ±c).
• Foci: (0, ±5√3)

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 10 = 20

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 5 = 10

Finding the Eccentricity:

• Eccentricity (e) = c / b = (5√3) / 10 = √3 / 2

Finding the Length of the Latus Rectum:

Length of latus rectum = 2a^2 / b = 2 * 25 / 10 = 5

5

Ans : 

Analyzing the Ellipse: x^2/49 + y^2/36 = 1

Standard Form of an Ellipse:

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 49
• b^2 = 36

Finding the Vertices:

• Since a^2 > b^2, the major axis is along the x-axis.
• The vertices are (±a, 0).
• a = √49 = 7
• Vertices: (±7, 0)

Finding the Foci:

• c = √(a^2 – b^2) = √(49 – 36) = √13
• The foci are (±c, 0).
• Foci: (±√13, 0)

Finding the Length of the Major Axis:

• Length of major axis = 2a 
• = 2 * 7 
• = 14

Finding the Length of the Minor Axis:

• Length of minor axis = 2b = 2 * 6 = 12

Finding the Eccentricity:

• Eccentricity (e) = c / a = (√13) / 7

Finding the Length of the Latus Rectum:

Length of latus rectum = 2b^2 / a

 = 2 * 36 / 7 

6.

Ans : 

(x^2 / a^2) + (y^2 / b^2) = 1

where:

• a^2 = 100
• b^2 = 400

Finding the Vertices:

• Since a^2 < b^2, the major axis is along the y-axis.
  
• The vertices are (0, ±b).
  
• b = √400 = 20
  
• Vertices: (0, ±20)
  

Finding the Foci:

• c = √(b^2 – a^2) = √(400 – 100) = √300 = 10√3
  
• The foci are (0, ±c).
  
• Foci: (0, ±10√3)
  

Finding the Length of the Major Axis:

• Length of major axis = 2b = 2 * 20 = 40

Finding the Length of the Minor Axis:

• Length of minor axis = 2a = 2 * 10 = 20

Finding the Eccentricity:

• Eccentricity (e) = c / b = (10√3) / 20 = √3 / 2

Finding the Length of the Latus Rectum:

• Length of latus rectum = 2a^2 / b = 2 * 100 / 20 = 10

7. 36x2 + 4y2 = 144 

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 144:

(36x^2 / 144) + (4y^2 / 144) = 144 / 144

Simplify:

x^2 / 4 + y^2 / 36 = 1

Step 2: Identify a^2 and b^2.

From the standard form, we can see that:

• a^2 = 4
• b^2 = 36

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(b^2 – a^2)

Substituting the values:

c = √(36 – 4) 

= √32 

= 4√2

Step 4: Determine the major and minor axes.

Since a^2 < b^2, the major axis is along the y-axis, and the minor axis is along the x-axis.

• Length of major axis = 2b 
• = 2 * √36 
• = 12
• Length of minor axis = 2a = 2 * √4 = 4

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (0, ±b) = (0, ±6)
• Foci: (0, ±c) = (0, ±4√2)
• Eccentricity (e): e = c / b = (4√2) / 6 = 2√2 / 3
• Length of latus rectum: 2a^2 / b = 2 * 4 / 6 = 4/3

8. 162+y2=16

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 16:

(x^2 / 1) + (y^2 / 16) = 1

Step 2: Identify a^2 and b^2.

• a^2 = 1
• b^2 = 16

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(b^2 – a^2)

Substituting the values:

c = √(16 – 1) = √15

Step 4: Determine the major and minor axes.

Since a^2 < b^2, the major axis is along the y-axis, and the minor axis is along the x-axis.

• Length of major axis = 2b = 2 * √16 = 8
• Length of minor axis = 2a = 2 * √1 = 2

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (0, ±b) = (0, ±4)
• Foci: (0, ±c) = (0, ±√15)
• Eccentricity (e): e = c / b = (√15) / 4
• Length of latus rectum: 2a^2 / b 
• = 2 * 1 / 4 
• = 1/2

9. 4x2 + 9y2 = 36

Ans : 

Step 1: Convert the equation to standard form.

To put the equation in standard form, divide both sides by 36:

(4x^2 / 36) + (9y^2 / 36) = 36 / 36

Simplify:

x^2 / 9 + y^2 / 4 = 1

Step 2: Identify a^2 and b^2.

• a^2 = 9
• b^2 = 4

Step 3: Calculate c.

The value of c (the distance from the center to the foci) is calculated using the formula:

c = √(a^2 – b^2)

Substituting the values:

c = √(9 – 4) = √5

Step 4: Determine the major and minor axes.

Since a^2 > b^2, the major axis is along the x-axis, and the minor axis is along the y-axis.

• Length of major axis = 2a = 2 * √9 = 6
• Length of minor axis = 2b = 2 * √4 = 4

Step 5: Find the coordinates of the foci, vertices, and other points.

• Vertices: (±a, 0) = (±3, 0)
• Foci: (±c, 0) = (±√5, 0)
• Eccentricity (e): e = c / a = (√5) / 3
• Length of latus rectum: 2b^2 / a = 2 * 4 / 3 = 8/3

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions: 

NCERT Solutions for Class 11 Maths Chapter 10

10. Vertices (± 5, 0), foci (± 4, 0) 

Ans : 

Given:

• Vertices: (±5, 0)
• Foci: (±4, 0)

From the vertices, we can determine:

• a = 5

From the foci, we can determine:

• c = 4

• c^2 = a^2 – b^2
• 16 = 25 – b^2
• b^2 = 9

(x^2 / 25) + (y^2 / 9) = 1

11. Vertices (0, ± 13), foci (0, ± 5)

Ans : 

Given:

• Vertices: (0, ±13)
• Foci: (0, ±5)

From the vertices, we can determine:

• b = 13

From the foci, we can determine:

• c = 5

• c^2 = b^2 – a^2
• 25 = 169 – a^2
• a^2 = 144

(x^2 / 144) + (y^2 / 169) = 1

Therefore, the equation of the ellipse is:

(x^2 / 144) + (y^2 / 169) = 1

12. Vertices (± 6, 0), foci (± 4, 0)

Ans : 

Given:

• Vertices: (±6, 0)
• Foci: (±4, 0)

From the vertices, we can determine:

• a = 6

From the foci, we can determine:

• c = 4

• c^2 = a^2 – b^2
• 16 = 36 – b^2
• b^2 = 20

(x^2 / 36) + (y^2 / 20) = 1

13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Ans :

Given:

• Ends of major axis: (±3, 0)
• Ends of minor axis: (0, ±2)

From the given points, we can determine:

• a = 3
• b = 2

(x^2 / 9) + (y^2 / 4) = 1

14. Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0)

Ans : 

15. Length of major axis 26, foci (± 5, 0)

Ans : 

Given:

• Length of major axis = 26
• Foci: (±5, 0)

• 2a = 26
• a = 13

From the foci, we can determine:

• c = 5

• c^2 = a^2 – b^2
• 25 = 169 – b^2
• b^2 = 144

(x^2 / 169) + (y^2 / 144) = 1

16. Length of minor axis 16, foci (0, ± 6). 

Ans : 

17. Foci (± 3, 0), a = 4

Ans : 

Given:

• Foci: (±3, 0)
• a = 4

From the foci, we can determine:

• c = 3  

• c^2 = a^2 – b^2
• 9 = 16 – b^2
• b^2 = 7  

(x^2 / 16) + (y^2 / 7) = 1

18. b = 3, c = 4, centre at the origin; foci on the x axis

Ans : 

Given:

• b = 3
• c = 4
• Center: (0, 0)

Since the center is at the origin, the standard form remains the same.

• c^2 = a^2 – b^2
• 16 = a^2 – 9
• a^2 = 25

(x^2 / 25) + (y^2 / 9) = 1

19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Ans : 

20. Major axis on the x-axis and passes through the points (4,3) and (6,2).

Ans :

NCERT Solutions for Class 11 Maths Chapter 10

FAQ’s

What is Class 11 Maths Chapter 10 about?

This chapter explains Conic Sections, including circles, ellipses, parabolas, and hyperbolas, along with their standard equations.

Why are Conic Sections important in Class 11 Maths?

Conic Sections form the base for advanced geometry and are used in fields like physics, astronomy, engineering, and architecture.

What are the main topics covered in Chapter 10 Conic Sections?

The chapter covers sections of a cone, focus–directrix property, standard forms of curves, and solving related problems.

How do Class 11 Maths Chapter 10 solutions help students?

They provide step-by-step explanations that make it easier to understand formulas, solve numerical problems, and prepare for exams.

Where can I find free solutions for Class 11 Maths Chapter 10?

You can get free solutions for Class 11 Maths Chapter 10 – Conic Sections – on trusted educational websites and learning platforms.