RD Sharma Solutions for Class 11 Maths Chapter 26 Ellipse

RD Sharma Solutions Class 11 Maths Chapter 26 – Avail Free PDF Updated for (2023-24)

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse is provided here for students to improve conceptual knowledge and score good marks in the board exams. In the previous section, we discussed the parabola. In this chapter, we shall study 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 board exam, we at BYJU’S have formulated the RD Sharma Class 11 Maths Solutions.

Chapter 26 – Ellipse contains one exercise, and RD Sharma Solutions provide comprehensive answers to all the questions present in this exercise. Students can easily download RD Sharma Solutions and can start practising offline in order to gain that extra edge of knowledge. To make learning fun and interesting for students, the links to this chapter are provided for free in PDF format, which is readily available to download from the links given below. 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.

RD Sharma Solutions for Class 11 Maths Chapter 26 – Ellipse

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Access 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

We also know that the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP² = e²PM²

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

We also know that the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP² = e²PM²

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

We also know that the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP² = e²PM²

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

We also know that the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP² = e²PM²

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

We also know that the perpendicular distance from the point (x₁, y₁) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP² = e²PM²

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

By using the formula,

Eccentricity:

Here, a² = ¼, b² = 1/9

Length of latus rectum = 2b²/a

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

= 4/9

Coordinates of foci (±ae, 0)

(ii) 5x² + 4y² = 1

Given:

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

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a² = 1/5 and b² = ¼

Length of latus rectum = 2b²/a

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

= 4/5

Coordinates of foci (±ae, 0)

(iii) 4x² + 3y² = 1

Given:

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

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a² = 1/4 and b² = 1/3

Length of latus rectum = 2b²/a

= [2(1/4)] / (1/√3)

= √3/2

Coordinates of foci (±ae, 0)

(iv) 25x² + 16y² = 1600

Given:

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

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a² = 64 and b² = 100

Length of latus rectum = 2b²/a

= [2(64)] / (100)

= 32/25

Coordinates of foci (±ae, 0)

(v) 9x² + 25y² = 225

Given:

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

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a² = 25 and b² = 9

Length of latus rectum = 2b²/a

= [2(9)] / (5)

= 18/5

Coordinates of foci (±ae, 0)

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
…. (1)

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

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:

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

By using the formula,

Eccentricity:

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

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

By using the formula,

Eccentricity:

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

So the equation of the ellipse can be given as

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

By using the formula,

Eccentricity:

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

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

By using the formula,

Eccentricity:

b² = 3a²/4

It is given that length of the 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

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
…. (1)

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

b² + 16a² = a² b² …. (2)

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

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),

So the equation of the ellipse can be given as

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

By using the formula,

Eccentricity:

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

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:

The 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

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

By using the formula,

Eccentricity:

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

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 the major axis is parallel to the 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

Since the major axis is parallel to the x-axis

So, a = 3 and b = 2.

a² = 9

b² = 4

So the equation of the ellipse can be given as

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 the major axis is parallel to the 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

Since the major axis is parallel to the y-axis

So, a = 2 and b = 3.

a² = 4

b² = 9

So the equation of the ellipse can be given as

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 the eccentricity of an ellipse is given as

(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 the eccentricity of an ellipse is given as