# RD Sharma Solutions for Class 11 Maths Chapter 26 Ellipse

## RD Sharma Solutions Class 11 Maths Chapter 26 â€“ Avail Free PDF Updated for (2021-22)

RD Sharma Solutions for Class 11 Maths Chapter 26 – EllipseÂ are provided here for students to score good marks in the board exams. 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 BYJUâ€™S have formulated the RD Sharma Class 11 Maths Solutions.

Students can easily download the RD Sharma 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

### 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 â€“ 2y + 5 = 0 and eccentricity equal to 1/2.

Solution:

Given:

Focus = (1, -2)

Directrix = 3x â€“ 2y + 5 = 0

Eccentricity = Â½

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

We know that distance between the points (x1, y1) and (x2, y2) isÂ 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

SP2Â = e2PM2

Upon cross multiplying, we get

52x2Â + 52y2Â – 104x + 208y + 260 = 9x2Â + 4y2Â – 12xy – 20y + 30x + 25

43x2Â + 48y2Â + 12xy – 134x + 228y + 235 = 0

âˆ´Â The equation of the ellipse is 43x2Â + 48y2Â + 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 = Â½.

(ii) focus is (- 1, 1), directrix is x – y + 3 = 0 and e = Â½.

(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 andÂ e = Â½.

Solution:

(i) focus is (0, 1), directrix is x + y = 0 and e = Â½

Given:

Focus is (0, 1)

Directrix is x + y = 0

e = Â½

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

We know that distance between the points (x1, y1) and (x2, y2) isÂ 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

SP2Â = e2PM2

Upon cross multiplying, we get

8x2Â + 8y2Â – 16y + 8 = x2Â + y2Â + 2xy

7x2Â + 7y2Â – 2xy – 16y + 8 = 0

âˆ´Â The equation of the ellipse is 7x2Â + 7y2Â – 2xy – 16y + 8 = 0

(ii) focus is (- 1, 1), directrix is x – y + 3 = 0 and e = Â½

Given:

Focus is (- 1, 1)

Directrix is x – y + 3 = 0

e = Â½

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

We know that distance between the points (x1, y1) and (x2, y2) isÂ 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

SP2Â = e2PM2

Upon cross multiplying, we get

8x2Â + 8y2Â + 16x – 16y + 16 = x2Â + y2Â – 2xy + 6x – 6y + 9

7x2Â + 7y2Â + 2xy + 10x – 10y + 7 = 0

âˆ´Â The equation of the ellipse is 7x2Â + 7y2Â + 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) isÂ 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

SP2Â = e2PM2

Upon cross multiplying, we get

325x2Â + 325y2Â + 1300x – 1950y + 4225 = 64x2Â + 144y2Â + 192xy + 256x + 384y + 256

261x2Â + 181y2Â – 192xy + 1044x – 2334y + 3969 = 0

âˆ´Â The equation of the ellipse is 261x2Â + 181y2Â – 192xy + 1044x – 2334y + 3969 = 0

(iv) focus is (1, 2), directrix is 3x + 4y – 7 = 0 andÂ e = Â½.

Given:

focus is (1, 2)

directrix is 3x + 4y – 7 = 0

e = Â½.

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

We know that distance between the points (x1, y1) and (x2, y2) isÂ 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

SP2Â = e2PM2

Upon cross multiplying, we get

100x2Â + 100y2Â – 200x – 400y + 500 = 9x2Â + 16y2Â + 24xy – 30x – 40y + 25

91x2Â + 84y2Â – 24xy – 170x – 360y + 475 = 0

âˆ´Â The equation of the ellipse is 91x2Â + 84y2Â – 24xy – 170x – 360y + 475 = 0

3. Find the eccentricity, coordinates of foci, length of the latus – rectum of the following ellipse:
(i) 4x2Â + 9y2Â = 1

(ii) 5x2Â + 4y2Â = 1

(iii) 4x2Â + 3y2Â = 1

(iv) 25x2Â + 16y2Â = 1600

(v) 9x2Â + 25y2Â = 225

Solution:

(i) 4x2Â + 9y2Â = 1

Given:

The equation of ellipse => 4x2Â + 9y2Â = 1

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a2 = Â¼, b2 = 1/9

Length of latus rectum = 2b2/a

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

= 4/9

Coordinates of foci (Â±ae, 0)

(ii) 5x2Â + 4y2Â = 1

Given:

The equation of ellipse => 5x2Â + 4y2Â = 1

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a2 = 1/5 and b2 = Â¼

Length of latus rectum = 2b2/a

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

= 4/5

Coordinates of foci (Â±ae, 0)

(iii) 4x2Â + 3y2Â = 1

Given:

The equation of ellipse => 4x2Â + 3y2Â = 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/âˆš3)

= âˆš3/2

Coordinates of foci (Â±ae, 0)

(iv) 25x2Â + 16y2Â = 1600

Given:

The equation of ellipse => 25x2Â + 16y2Â = 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 (Â±ae, 0)

(v) 9x2Â + 25y2Â = 225

Given:

The equation of ellipse => 9x2Â + 25y2Â = 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 (Â±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 âˆš(2/5).

Solution:

Given:

The point (-3, 1)

Eccentricity = âˆš(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
â€¦. (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)2Â + 5(1)2Â = 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:

3x2Â + 5y2Â = 32

âˆ´Â The equation of the ellipse is 3x2Â + 5y2Â = 32.

5. Find the equation of the ellipse in the following cases:
(i) eccentricity e = Â½Â and foci (Â± 2, 0)

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

(iii) eccentricity e = Â½Â and semi – major axis = 4

(iv) eccentricity e = Â½ and major axis = 12

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

Solution:

(i) Eccentricity e = Â½Â and foci (Â± 2, 0)

Given:

Eccentricity e = Â½

Foci (Â± 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 (Â± 2, 0) =>foci = (Â±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

3x2Â + 4y2Â = 48

âˆ´Â The equation of the ellipse is 3x2Â + 4y2Â = 48

(ii) eccentricity e = 2/3Â 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

20x2Â + 36y2Â = 405

âˆ´Â The equation of the ellipse is 20x2Â + 36y2Â = 405.

(iii) eccentricity e = Â½Â and semi – major axis = 4

Given:

Eccentricity e = Â½

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

a2Â = 16

We know, b2 = 3a2/4

b2 = 3(16)/4

= 4

So the equation of the ellipse can be given as

3x2Â + 4y2Â = 48

âˆ´Â The equation of the ellipse is 3x2Â + 4y2Â = 48.

(iv) eccentricity e = Â½ and major axis = 12

Given:

Eccentricity e = Â½

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 major axis is 2a.

2a = 12

a = 6

a2Â = 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

3x2Â + 4y2Â = 108

âˆ´Â The equation of the ellipse is 3x2Â + 4y2Â = 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
â€¦. (1)

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

b2Â + 16a2Â = a2Â b2 â€¦. (2)

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

a2Â + 36b2Â = a2b2 â€¦. (3)

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

(16a2Â + 576b2) – (b2Â + 16a2) = (16a2b2Â – a2b2)

575b2Â = 15a2b2

15a2Â = 575

a2 = 575/15

= 115/3

So from equation (2),

So the equation of the ellipse can be given as

3x2Â + 7y2Â = 115

âˆ´Â The equation of the ellipse is 3x2Â + 7y2Â = 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 = (Â±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:

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 length of the minor axis is 2b and 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

a2Â = 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

x2Â + 2y2Â = 100

âˆ´Â The equation of the ellipse is x2Â + 2y2Â = 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

Since major axis is parallel to x – axis

So, a = 3 and b = 2.

a2Â = 9

b2Â = 4

So the equation of the ellipse can be given as

4(x2Â + 4x + 4) + 9(y2Â – 6y + 9) = 36

4x2Â + 16x + 16 + 9y2Â – 54y + 81 = 36

4x2Â + 9y2Â + 16x – 54y + 61 = 0

âˆ´Â The equation of the ellipse is 4x2Â + 9y2Â + 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

Since major axis is parallel to y – axis

So, a = 2 and b = 3.

a2Â = 4

b2Â = 9

So the equation of the ellipse can be given as

9(x2Â + 4x + 4) + 4(y2Â – 6y + 9) = 36

9x2Â + 36x + 36 + 4y2Â – 24y + 36 = 36

9x2Â + 4y2Â + 36x – 24y + 36 = 0

âˆ´Â The equation of the ellipse is 9x2Â + 4y2Â + 36x – 24y + 36 = 0.

9. Find the eccentricity of an ellipse whose latus – rectum is
(i)Â Half of its minor axis
(ii)Â 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 isÂ 2b2/a.

2b2/a = b

a = 2b â€¦. (1)

By using the formula,

We know that 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 â€¦. (1)

By using the formula,

We know that eccentricity of an ellipse is given as