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

Class 11: Maths Chapter 26 solutions. Complete Class 11 Maths Chapter 26 Notes.

RD Sharma Solutions for Class 11 Maths Chapter 26–Ellipse

RD Sharma 11th Maths Chapter 26, Class 11 Maths Chapter 26 solutions

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 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:

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

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 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.

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 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.

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 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 eccentricity of an ellipse is given as

RD Sharma Solutions for Class 11 Maths Chapter 26: Download PDF

RD Sharma Solutions for Class 11 Maths Chapter 26–Ellipse

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About RD Sharma

RD Sharma isn’t the kind of author you’d bump into at lit fests. But his bestselling books have helped many CBSE students lose their dread of maths. Sunday Times profiles the tutor turned internet star
He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like ‘series solution of linear differential equations’. Meet Dr Ravi Dutt Sharma — mathematics teacher and author of 25 reference books — whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it’s only recently that a spoof video turned the tutor into a YouTube star.

R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. “I like to spend all my time thinking and writing about maths problems. I find it relaxing,” he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government’s Guru Nanak Dev Institute of Technology.

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