- RD Chapter 1- Sets
- RD Chapter 2- Relations
- RD Chapter 3- Functions
- RD Chapter 4- Measurement of Angles
- RD Chapter 5- Trigonometric Functions
- RD Chapter 6- Graphs of Trigonometric Functions
- RD Chapter 7- Trigonometric Ratios of Compound Angles
- RD Chapter 8- Transformation Formulae
- RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles
- RD Chapter 10- Sine and Cosine Formulae and Their Applications
- RD Chapter 11- Trigonometric Equations
- RD Chapter 12- Mathematical Induction
- RD Chapter 13- Complex Numbers
- RD Chapter 14- Quadratic Equations
- RD Chapter 15- Linear Inequations
- RD Chapter 16- Permutations
- RD Chapter 17- Combinations
- RD Chapter 18- Binomial Theorem
- RD Chapter 19- Arithmetic Progressions
- RD Chapter 20- Geometric Progressions
- RD Chapter 21- Some Special Series
- RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates
- RD Chapter 23- The Straight Lines
- RD Chapter 24- The Circle
- RD Chapter 25- Parabola
- RD Chapter 27- Hyperbola
- RD Chapter 28- Introduction to 3D coordinate geometry
- RD Chapter 29- Limits
- RD Chapter 30- Derivatives
- RD Chapter 31- Mathematical Reasoning
- RD Chapter 32- Statistics
- RD Chapter 33- Probability

RD Chapter 1- Sets |
RD Chapter 2- Relations |
RD Chapter 3- Functions |
RD Chapter 4- Measurement of Angles |
RD Chapter 5- Trigonometric Functions |
RD Chapter 6- Graphs of Trigonometric Functions |
RD Chapter 7- Trigonometric Ratios of Compound Angles |
RD Chapter 8- Transformation Formulae |
RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles |
RD Chapter 10- Sine and Cosine Formulae and Their Applications |
RD Chapter 11- Trigonometric Equations |
RD Chapter 12- Mathematical Induction |
RD Chapter 13- Complex Numbers |
RD Chapter 14- Quadratic Equations |
RD Chapter 15- Linear Inequations |
RD Chapter 16- Permutations |
RD Chapter 17- Combinations |
RD Chapter 18- Binomial Theorem |
RD Chapter 19- Arithmetic Progressions |
RD Chapter 20- Geometric Progressions |
RD Chapter 21- Some Special Series |
RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates |
RD Chapter 23- The Straight Lines |
RD Chapter 24- The Circle |
RD Chapter 25- Parabola |
RD Chapter 27- Hyperbola |
RD Chapter 28- Introduction to 3D coordinate geometry |
RD Chapter 29- Limits |
RD Chapter 30- Derivatives |
RD Chapter 31- Mathematical Reasoning |
RD Chapter 32- Statistics |
RD Chapter 33- Probability |

**Answer
1** :

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 (x_{1}, y_{1})and (x_{2}, y_{2}) is given as

We also know that the perpendicular distance from the point(x_{1}, y_{1}) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP^{2} = e^{2}PM^{2}

Upon cross multiplying, we get

52x^{2} + 52y^{2} – 104x + 208y +260 = 9x^{2} + 4y^{2} – 12xy – 20y + 30x + 25

43x^{2} + 48y^{2} + 12xy – 134x +228y + 235 = 0

∴ The equation of the ellipse is 43x^{2} +48y^{2} + 12xy – 134x + 228y + 235 = 0

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.

**Answer
2** :

(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 (x_{1}, y_{1})and (x_{2}, y_{2}) is given as

We also know that the perpendicular distance from the point(x_{1}, y_{1}) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP^{2} = e^{2}PM^{2}

Upon cross multiplying, we get

8x^{2} + 8y^{2} – 16y + 8 = x^{2} +y^{2} + 2xy

7x^{2} + 7y^{2} – 2xy – 16y + 8 =0

∴ The equation of the ellipse is 7x^{2} +7y^{2} – 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 (x_{1}, y_{1})and (x_{2}, y_{2}) is given as

We also know that the perpendicular distance from the point(x_{1}, y_{1}) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP^{2} = e^{2}PM^{2}

Upon cross multiplying, we get

8x^{2} + 8y^{2} + 16x – 16y + 16 =x^{2} + y^{2} – 2xy + 6x – 6y + 9

7x^{2} + 7y^{2} + 2xy + 10x – 10y+ 7 = 0

∴ The equation of the ellipse is 7x^{2} +7y^{2} + 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 (x_{1}, y_{1})and (x_{2}, y_{2}) is given as

_{1}, y_{1}) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP^{2} = e^{2}PM^{2}

Upon cross multiplying, we get

325x^{2} + 325y^{2} + 1300x –1950y + 4225 = 64x^{2} + 144y^{2} + 192xy + 256x +384y + 256

261x^{2} + 181y^{2} – 192xy +1044x – 2334y + 3969 = 0

∴ The equation of the ellipse is 261x^{2} +181y^{2} – 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 (x_{1}, y_{1})and (x_{2}, y_{2}) is given as

_{1}, y_{1}) to the line ax + by + c = 0 is given as

So,

SP = ePM

SP^{2} = e^{2}PM^{2}

Upon cross multiplying, we get

100x^{2} + 100y^{2} – 200x – 400y+ 500 = 9x^{2} + 16y^{2} + 24xy – 30x – 40y + 25

91x^{2} + 84y^{2} – 24xy – 170x –360y + 475 = 0

∴ The equation of the ellipse is 91x^{2} +84y^{2} – 24xy – 170x – 360y + 475 = 0

Find the eccentricity, coordinatesof foci, length of the latus – rectum of the following ellipse:

(i) 4x^{2} + 9y^{2} = 1

(ii) 5x^{2} + 4y^{2} =1

(iii) 4x^{2} + 3y^{2} =1

(iv) 25x^{2} + 16y^{2} =1600

(v) 9x^{2} + 25y^{2} =225

**Answer
3** :

(i) 4x^{2} + 9y^{2} = 1

Given:

The equation of ellipse => 4x^{2} + 9y^{2} =1

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a^{2} = ¼, b^{2} = 1/9

Length of latus rectum = 2b^{2}/a

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

= 4/9

Coordinates of foci (±ae, 0)

(ii) 5x^{2} + 4y^{2} = 1

Given:

The equation of ellipse => 5x^{2} + 4y^{2} =1

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a^{2} = 1/5 and b^{2} = ¼

Length of latus rectum = 2b^{2}/a

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

= 4/5

Coordinates of foci (±ae, 0)

(iii) 4x^{2} + 3y^{2} = 1

Given:

The equation of ellipse => 4x^{2} + 3y^{2} =1

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a^{2} = 1/4 and b^{2} = 1/3

Length of latus rectum = 2b^{2}/a

= [2(1/4)] / (1/√3)

= √3/2

Coordinates of foci (±ae, 0)

(iv) 25x^{2} + 16y^{2} = 1600

Given:

The equation of ellipse => 25x^{2} + 16y^{2} =1600

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a^{2} = 64 and b^{2} = 100

Length of latus rectum = 2b^{2}/a

= [2(64)] / (100)

= 32/25

Coordinates of foci (±ae, 0)

(v) 9x^{2} + 25y^{2} = 225

Given:

The equation of ellipse => 9x^{2} + 25y^{2} =225

This equation can be expressed as

By using the formula,

Eccentricity:

Here, a^{2} = 25 and b^{2} = 9

Length of latus rectum = 2b^{2}/a

= [2(9)] / (5)

= 18/5

Coordinates of foci (±ae, 0)

**Answer
4** :

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 xand 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} = 3a^{2}

3(9) + 5 = 3a^{2}

32 = 3a^{2}

a^{2} = 32/3

From equation (2)

b^{2} = 3a^{2}/5

= 3(32/3) / 5

= 32/5

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

3x^{2} + 5y^{2} = 32

∴ The equation of the ellipse is 3x^{2} +5y^{2} = 32.

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)

**Answer
5** :

(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 xand y – axis is given as

By using the formula,

Eccentricity:

b^{2} = 3a^{2}/4

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

Where, ae = 2

a(1/2) = 2

a = 4

a^{2} = 16

We know b^{2} = 3a^{2}/4

b^{2} = 3(16)/4

= 12

So the equation of the ellipse can be given as

3x^{2} + 4y^{2} = 48

∴ The equation of the ellipse is 3x^{2} +4y^{2} = 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 xand y – axis is given as

By using the formula,

Eccentricity:

By using the formula, length of the latus rectum is 2b^{2}/a

So the equation of the ellipse can be given as

20x^{2} + 36y^{2} = 405

∴ The equation of the ellipse is 20x^{2} +36y^{2} = 405.

(iii) eccentricity e = ½ andsemi – 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 xand 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

a^{2} = 16

We know, b^{2} = 3a^{2}/4

b^{2} = 3(16)/4

= 4

So the equation of the ellipse can be given as

3x^{2} + 4y^{2} = 48

∴ The equation of the ellipse is 3x^{2} +4y^{2} = 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 xand y – axis is given as

By using the formula,

Eccentricity:

b^{2} = 3a^{2}/4

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

2a = 12

a = 6

a^{2} = 36

So, by substituting the value of a^{2}, we get

b^{2} = 3(36)/4

= 27

So the equation of the ellipse can be given as

3x^{2} + 4y^{2} = 108

∴ The equation of the ellipse is 3x^{2} +4y^{2} = 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 xand y – axis is given as

…. (1)

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

b^{2} + 16a^{2} = a^{2} b^{2} ….(2)

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

a^{2} + 36b^{2} = a^{2}b^{2} ….(3)

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

(16a^{2} + 576b^{2}) – (b^{2} +16a^{2}) = (16a^{2}b^{2} – a^{2}b^{2})

575b^{2} = 15a^{2}b^{2}

15a^{2} = 575

a^{2} = 575/15

= 115/3

So from equation (2),

So the equation of the ellipse can be given as

3x^{2} + 7y^{2} = 115

∴ The equation of the ellipse is 3x^{2} +7y^{2} = 115.

**Answer
6** :

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 xand 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

a^{2} = 144

By substituting the value of a^{2}, we get

b^{2} = 8a^{2}/9

b^{2} = 8(144)/9

= 128

So the equation of the ellipse can be given as

**Answer
7** :

Given:

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

Now let us find the equation to the ellipse.

We know that the equation of the ellipse whose axes are xand y – axis is given as

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

By using the formula,

Eccentricity:

We know that the length of the latus rectum is 2b^{2}/a

It is given that length of the latus rectum = 10

So by equating, we get

2b^{2}/a = 10

a^{2}/ a = 10 [Since, a^{2} = 2b^{2}]

a = 10

a^{2} = 100

Now, by substituting the value of a^{2} we get

2b^{2}/a = 10

2b^{2}/10 = 10

2b^{2} = 10(10)

b^{2} = 100/2

= 50

So the equation of the ellipse can be given as

x^{2} + 2y^{2} = 100

∴ The equation of the ellipse is x^{2} + 2y^{2} =100.

**Answer
8** :

Given:

Centre = (-2, 3)

Semi – axis are 3 and 2

(i) Whenmajor 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.

a^{2} = 9

b^{2} = 4

So the equation of the ellipse can be given as

4(x^{2} + 4x + 4) + 9(y^{2} – 6y +9) = 36

4x^{2} + 16x + 16 + 9y^{2} – 54y +81 = 36

4x^{2} + 9y^{2} + 16x – 54y + 61 =0

∴ The equation of the ellipse is 4x^{2} +9y^{2} + 16x – 54y + 61 = 0.

(ii) Whenmajor 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.

a^{2} = 4

b^{2} = 9

So the equation of the ellipse can be given as

9(x^{2} + 4x + 4) + 4(y^{2} – 6y +9) = 36

9x^{2} + 36x + 36 + 4y^{2} – 24y +36 = 36

9x^{2} + 4y^{2} + 36x – 24y + 36 =0

∴ The equation of the ellipse is 9x^{2} +4y^{2} + 36x – 24y + 36 = 0.

Find the eccentricity of an ellipse whose latus – rectum is

(i) Half of its minor axis

(ii) Half of its major axis

**Answer
9** :

Given:

We need to find the eccentricity of an ellipse.

(i) Iflatus – rectum is half of its minor axis

We know that the length of the semi – minor axis is b andthe length of the latus – rectum is 2b^{2}/a.

2b^{2}/a = b

a = 2b …. (1)

By using the formula,

We know that eccentricity of an ellipse is given as

(ii) Iflatus – rectum is half of its major axis

We know that the length of the semi – major axis is a andthe length of the latus – rectum is 2b^{2}/a.

2b^{2}/a

a^{2} = 2b^{2} …. (1)

By using the formula,

We know that eccentricity of an ellipse is given as

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