RD Sharma Solutions for Class 11 Maths Chapter 26 Ellipse

In the previous section, we have discussed parabola. In this chapter we shall study about the ellipse and also find the equation of an ellipse in standard and other forms. In order to enhance the performance of students in the Class 11 exam, we at BYJU’S have formulated the RD Sharma Class 11 Maths Solutions. Students can easily download the solutions and can start practising offline in order to gain that extra edge of knowledge. Students are also advised to practice on a daily basis to excel in the board exams. To help gain interest among students, the links to this chapter are provided for free in the pdf format, which is readily available to download from the below-mentioned links.

Chapter 26 – Ellipse contains one exercise and the RD Sharma Solutions present in this page provide solutions to the questions present in this exercise. Now, let us have a look at the concepts discussed in this chapter.

  • Equation of the ellipse in standard form.
    • Tracing of the ellipse.
    • Second focus and second directrix of the ellipse.
    • Vertices, major and minor axes, foci, directrices and centre.
    • Ordinate, double ordinate and latus-rectum.
    • Focal distances of a point on the ellipse.
  • Equation of the ellipse in other forms.

Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse

 

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Acccess RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse

EXERCISE 26.1 PAGE NO: 26.22

1. Find the equation of the ellipse whose focus is (1, -2), the directrix 3x – 2y + 5 = 0 and eccentricity equal to 1/2.

Solution:

Given:

Focus = (1, -2)

Directrix = 3x – 2y + 5 = 0

Eccentricity = ½

Let P(x, y) be any point on the ellipse.

We know that distance between the points (x1, y1) and (x2, y2) is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 1

We also know that the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is given as 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 2

So,

SP = ePM

SP2 = e2PM2

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 3

Upon cross multiplying, we get

52x2 + 52y2 – 104x + 208y + 260 = 9x2 + 4y2 – 12xy – 20y + 30x + 25

43x2 + 48y2 + 12xy – 134x + 228y + 235 = 0

∴ The equation of the ellipse is 43x2 + 48y2 + 12xy – 134x + 228y + 235 = 0

2. Find the equation of the ellipse in the following cases:

(i) focus is (0, 1), directrix is x + y = 0 and e = ½.

(ii) focus is (- 1, 1), directrix is x – y + 3 = 0 and e = ½.

(iii) focus is (- 2, 3), directrix is 2x + 3y + 4 = 0 and e = 4/5.

(iv) focus is (1, 2), directrix is 3x + 4y – 7 = 0 and e = ½.

Solution:

(i) focus is (0, 1), directrix is x + y = 0 and e = ½

Given:

Focus is (0, 1)

Directrix is x + y = 0

e = ½

Let P(x, y) be any point on the ellipse.

We know that distance between the points (x1, y1) and (x2, y2) is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 4

We also know that the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is given as 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 5

So,

SP = ePM

SP2 = e2PM2

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 6

Upon cross multiplying, we get

8x2 + 8y2 – 16y + 8 = x2 + y2 + 2xy

7x2 + 7y2 – 2xy – 16y + 8 = 0

∴ The equation of the ellipse is 7x2 + 7y2 – 2xy – 16y + 8 = 0

(ii) focus is (- 1, 1), directrix is x – y + 3 = 0 and e = ½

Given:

Focus is (- 1, 1)

Directrix is x – y + 3 = 0

e = ½

Let P(x, y) be any point on the ellipse.

We know that distance between the points (x1, y1) and (x2, y2) is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 7

We also know that the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is given as 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 8

So,

SP = ePM

SP2 = e2PM2

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 9

Upon cross multiplying, we get

 8x2 + 8y2 + 16x – 16y + 16 = x2 + y2 – 2xy + 6x – 6y + 9

7x2 + 7y2 + 2xy + 10x – 10y + 7 = 0

∴ The equation of the ellipse is 7x2 + 7y2 + 2xy + 10x – 10y + 7 = 0

(iii) focus is (- 2, 3), directrix is 2x + 3y + 4 = 0 and e = 4/5

Focus is (- 2, 3)

Directrix is 2x + 3y + 4 = 0

e = 4/5

Let P(x, y) be any point on the ellipse.

We know that distance between the points (x1, y1) and (x2, y2) is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 10

We also know that the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is given as 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 11

So,

SP = ePM

SP2 = e2PM2

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 12

Upon cross multiplying, we get

325x2 + 325y2 + 1300x – 1950y + 4225 = 64x2 + 144y2 + 192xy + 256x + 384y + 256

261x2 + 181y2 – 192xy + 1044x – 2334y + 3969 = 0

∴ The equation of the ellipse is 261x2 + 181y2 – 192xy + 1044x – 2334y + 3969 = 0

(iv) focus is (1, 2), directrix is 3x + 4y – 7 = 0 and e = ½.

Given:

focus is (1, 2)

directrix is 3x + 4y – 7 = 0

e = ½.

Let P(x, y) be any point on the ellipse.

We know that distance between the points (x1, y1) and (x2, y2) is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 13

We also know that the perpendicular distance from the point (x1, y1) to the line ax + by + c = 0 is given as 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 14

So,

SP = ePM

SP2 = e2PM2

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 15

Upon cross multiplying, we get

100x2 + 100y2 – 200x – 400y + 500 = 9x2 + 16y2 + 24xy – 30x – 40y + 25

91x2 + 84y2 – 24xy – 170x – 360y + 475 = 0

∴ The equation of the ellipse is 91x2 + 84y2 – 24xy – 170x – 360y + 475 = 0

3. Find the eccentricity, coordinates of foci, length of the latus – rectum of the following ellipse:
(i) 4x2 + 9y2 = 1

(ii) 5x2 + 4y2 = 1

(iii) 4x2 + 3y2 = 1

(iv) 25x2 + 16y2 = 1600

(v) 9x2 + 25y2 = 225

Solution:

(i) 4x2 + 9y2 = 1

Given:

The equation of ellipse => 4x2 + 9y2 = 1

This equation can be expressed as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 16

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 17

Here, a2 = ¼, b2 = 1/9

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 18

Length of latus rectum = 2b2/a

= [2 (1/9)] / (1/2)

= 4/9

 Coordinates of foci (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 19

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 20

(ii) 5x2 + 4y2 = 1

Given:

The equation of ellipse => 5x2 + 4y2 = 1

This equation can be expressed as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 21

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 22

Here, a2 = 1/5 and b2 = ¼

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 23

Length of latus rectum = 2b2/a

= [2(1/5)] / (1/2)

= 4/5

Coordinates of foci (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 24

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 25

(iii) 4x2 + 3y2 = 1

Given:

The equation of ellipse => 4x2 + 3y2 = 1

This equation can be expressed as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 26

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 27

Here, a2 = 1/4 and b2 = 1/3

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 28

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 29

Length of latus rectum = 2b2/a

= [2(1/4)] / (1/3)

= 3/2

Coordinates of foci (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 30

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 31

(iv) 25x2 + 16y2 = 1600

Given:

The equation of ellipse => 25x2 + 16y2 = 1600

This equation can be expressed as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 32

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 33

Here, a2 = 64 and b2 = 100

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 34

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 35

Length of latus rectum = 2b2/a

= [2(64)] / (100)

= 32/25

Coordinates of foci (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 36
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 37

(v) 9x2 + 25y2 = 225

Given:

The equation of ellipse => 9x2 + 25y2 = 225

This equation can be expressed as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 38

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 39

Here, a2 = 25 and b2 = 9

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 40

Length of latus rectum = 2b2/a

= [2(9)] / (5)

= 18/5

Coordinates of foci (±ae, 0)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 41

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 42

4. Find the equation to the ellipse (referred to its axes as the axes of x and y respectively) which passes through the point (-3, 1) and has eccentricity √(2/5).

Solution:

Given:

The point (-3, 1)

Eccentricity = √(2/5)

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 43…. (1)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 44

Now let us substitute equation (2) in equation (1), we get

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 45

It is given that the curve passes through the point (-3, 1).

So by substituting the point in the curve we get,

3(- 3)2 + 5(1)2 = 3a2

3(9) + 5 = 3a2

32 = 3a2

a2 = 32/3

From equation (2)

b2 = 3a2/5

= 3(32/3) / 5

= 32/5

So now, the equation of the ellipse is given as:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 46

3x2 + 5y2 = 32

∴ The equation of the ellipse is 3x2 + 5y2 = 32.

5. Find the equation of the ellipse in the following cases:
(i) eccentricity e = ½ and foci (± 2, 0)

(ii) eccentricity e = 2/3 and length of latus – rectum = 5

(iii) eccentricity e = ½ and semi – major axis = 4

(iv) eccentricity e = ½ and major axis = 12

(v) The ellipse passes through (1, 4) and (- 6, 1)

Solution:

(i) Eccentricity e = ½ and foci (± 2, 0)

Given:

Eccentricity e = ½ 

Foci (± 2, 0)

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 47

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 48

b2 = 3a2/4

It is given that foci (± 2, 0) =>foci = (±ae, 0)

Where, ae = 2

a(1/2) = 2

a = 4

a2 = 16

We know b2 = 3a2/4

b2 = 3(16)/4

= 12

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 49

3x2 + 4y2 = 48

∴ The equation of the ellipse is 3x2 + 4y2 = 48

(ii) eccentricity e = 2/3 and length of latus rectum = 5

Given:

Eccentricity e = 2/3

Length of latus – rectum = 5

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 50

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 51

By using the formula, length of the latus rectum is 2b2/a

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 52

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 53

20x2 + 36y2 = 405

∴ The equation of the ellipse is 20x2 + 36y2 = 405.

(iii) eccentricity e = ½ and semi – major axis = 4

Given:

Eccentricity e = ½

Semi – major axis = 4

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 54

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 55

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 56

It is given that the length of the semi – major axis is a

a = 4

a2 = 16

We know, b2 = 3a2/4

b2 = 3(16)/4

= 4

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 57

3x2 + 4y2 = 48

∴ The equation of the ellipse is 3x2 + 4y2 = 48.

(iv) eccentricity e = ½ and major axis = 12

Given:

Eccentricity e = ½

Major axis = 12

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 58

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 59

b2 = 3a2/4

It is given that length of major axis is 2a.

2a = 12

a = 6

a2 = 36

So, by substituting the value of a2, we get

b2 = 3(36)/4

= 27

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 60

 3x2 + 4y2 = 108

∴ The equation of the ellipse is 3x2 + 4y2 = 108.

(v) The ellipse passes through (1, 4) and (- 6, 1)

Given:

The points (1, 4) and (- 6, 1)

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 61…. (1)

Let us substitute the point (1, 4) in equation (1), we get

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 62

b2 + 16a2 = a2 b2 …. (2)

Let us substitute the point (-6, 1) in equation (1), we get

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 63

a2 + 36b2 = a2b2 …. (3)

Let us multiply equation (3) by 16 and subtract with equation (2), we get

(16a2 + 576b2) – (b2 + 16a2) = (16a2b2 – a2b2)

575b2 = 15a2b2

15a2 = 575

a2 = 575/15

= 115/3

So from equation (2),

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 64

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 65

3x2 + 7y2 = 115

∴ The equation of the ellipse is 3x2 + 7y2 = 115.

6. Find the equation of the ellipse whose foci are (4, 0) and (- 4, 0), eccentricity = 1/3.

Solution:

Given:

Foci are (4, 0) (- 4, 0)

Eccentricity = 1/3.

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 66

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 67

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 68

It is given that foci = (4, 0) (- 4, 0) => foci = (±ae,0)

Where, ae = 4

a(1/3) = 4

a = 12

a2 = 144

By substituting the value of a2, we get

b2 = 8a2/9

b2 = 8(144)/9

= 128

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 69

7. Find the equation of the ellipse in the standard form whose minor axis is equal to the distance between foci and whose latus – rectum is 10.

Solution:

Given:

Minor axis is equal to the distance between foci and whose latus – rectum is 10.

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are x and y – axis is given as
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 70

We know that length of the minor axis is 2b and distance between the foci is 2ae.

By using the formula,

Eccentricity:

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 71

We know that the length of the latus rectum is 2b2/a

It is given that length of the latus rectum = 10

So by equating, we get

2b2/a = 10

a2/ a = 10 [Since, a2 = 2b2]

 a = 10

a2 = 100

Now, by substituting the value of a2 we get

2b2/a = 10

2b2/10 = 10

2b2 = 10(10)

b2 = 100/2

= 50

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 72

x2 + 2y2 = 100

∴ The equation of the ellipse is x2 + 2y2 = 100.

8. Find the equation of the ellipse whose centre is (-2, 3) and whose semi – axis are 3 and 2 when the major axis is (i) parallel to x – axis (ii) parallel to the y – axis.

Solution:

Given:

Centre = (-2, 3)

Semi – axis are 3 and 2

(i) When major axis is parallel to x-axis

Now let us find the equation to the ellipse.

We know that the equation of the ellipse with centre (p, q) is given by 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 73

Since major axis is parallel to x – axis

So, a = 3 and b = 2.

a2 = 9

b2 = 4

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 74

4(x2 + 4x + 4) + 9(y2 – 6y + 9) = 36

4x2 + 16x + 16 + 9y2 – 54y + 81 = 36

4x2 + 9y2 + 16x – 54y + 61 = 0

∴ The equation of the ellipse is 4x2 + 9y2 + 16x – 54y + 61 = 0.

(ii) When major axis is parallel to y-axis

Now let us find the equation to the ellipse.

We know that the equation of the ellipse with centre (p, q) is given by 
RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 75

Since major axis is parallel to y – axis

So, a = 2 and b = 3.

a2 = 4

b2 = 9

So the equation of the ellipse can be given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 76

9(x2 + 4x + 4) + 4(y2 – 6y + 9) = 36

9x2 + 36x + 36 + 4y2 – 24y + 36 = 36

9x2 + 4y2 + 36x – 24y + 36 = 0

∴ The equation of the ellipse is 9x2 + 4y2 + 36x – 24y + 36 = 0.

9. Find the eccentricity of an ellipse whose latus – rectum is
(i) Half of its minor axis
(ii) Half of its major axis

Solution:

Given:

We need to find the eccentricity of an ellipse.


(i) If latus – rectum is half of its minor axis

We know that the length of the semi – minor axis is b and the length of the latus – rectum is 2b2/a.

2b2/a = b

a = 2b …. (1)

By using the formula,

We know that eccentricity of an ellipse is given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 77

(ii) If latus – rectum is half of its major axis

We know that the length of the semi – major axis is a and the length of the latus – rectum is 2b2/a.

2b2/a

a2 = 2b2 …. (1)

By using the formula,

We know that eccentricity of an ellipse is given as

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse - image 78

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