- 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 26- Ellipse
- 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 26- Ellipse |
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:

The equation of the directrix of a hyperbola => x – y + 3= 0.

Focus = (-1, 1) and

Eccentricity = 3

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]

So,

[We know that (a – b)^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

So, 2{x^{2} + 1 + 2x + y^{2} + 1 –2y} = 9{x^{2} + y^{2}+ 9 + 6x – 6y – 2xy}

2x^{2} + 2 + 4x + 2y^{2} + 2 – 4y= 9x^{2} + 9y^{2}+ 81 + 54x – 54y – 18xy

2x^{2} + 4 + 4x + 2y^{2}– 4y – 9x^{2} –9y^{2} – 81 – 54x + 54y + 18xy = 0

– 7x^{2} – 7y^{2} – 50x + 50y +18xy – 77 = 0

7(x^{2} + y^{2}) – 18xy + 50x – 50y +77 = 0

∴The equation of hyperbola is 7(x^{2} + y^{2})– 18xy + 50x – 50y + 77 = 0

Find the equation of the hyperbola whose

(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity = 2

(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity = 2

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3

(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity = 2

(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity = 2

(vi)focus is (2, 2), directrix is x + y = 9 and eccentricity = 2

**Answer
2** :

(i) focus is (0, 3), directrix is x + y – 1 = 0 and eccentricity= 2

Given:

Focus = (0, 3)

Directrix => x + y – 1 = 0

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]

So,

[We know that (a – b)^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

So, 2{x^{2} + y^{2} + 9 – 6y} =4{x^{2} + y^{2} + 1 – 2x – 2y + 2xy}

2x^{2} + 2y^{2} + 18 – 12y = 4x^{2} +4y^{2}+ 4 – 8x – 8y + 8xy

2x^{2} + 2y^{2} + 18 – 12y – 4x^{2} –4y^{2} – 4 – 8x + 8y – 8xy = 0

– 2x^{2} – 2y^{2} – 8x – 4y – 8xy+ 14 = 0

-2(x^{2} + y^{2} – 4x + 2y + 4xy –7) = 0

x^{2} + y^{2} – 4x + 2y + 4xy – 7= 0

∴The equation of hyperbola is x^{2} + y^{2} –4x + 2y + 4xy – 7 = 0

(ii) focus is (1, 1), directrix is 3x + 4y + 8 = 0 andeccentricity = 2

Focus = (1, 1)

Directrix => 3x + 4y + 8 = 0

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]

So,

[We know that (a – b)^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

25{x^{2} + 1 – 2x + y^{2} + 1 –2y} = 4{9x^{2} + 16y^{2}+ 64 + 24xy + 64y + 48x}

25x^{2} + 25 – 50x + 25y^{2} + 25– 50y = 36x^{2} + 64y^{2} + 256 + 96xy + 256y + 192x

25x^{2} + 25 – 50x + 25y^{2} + 25– 50y – 36x^{2} – 64y^{2} – 256 – 96xy – 256y –192x = 0

– 11x^{2} – 39y^{2} – 242x – 306y– 96xy – 206 = 0

11x^{2} + 96xy + 39y^{2} + 242x +306y + 206 = 0

∴The equation of hyperbola is11x^{2} + 96xy +39y^{2} + 242x + 306y + 206 = 0

(iii) focus is (1, 1) directrix is 2x + y = 1 and eccentricity =√3

Given:

Focus = (1, 1)

Directrix => 2x + y = 1

Eccentricity =√3

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

So,

^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

5{x^{2} + 1 – 2x + y^{2} + 1 – 2y}= 3{4x^{2} + y^{2}+ 1 + 4xy – 2y – 4x}

5x^{2} + 5 – 10x + 5y^{2} + 5 –10y = 12x^{2} + 3y^{2} + 3 + 12xy – 6y – 12x

5x^{2} + 5 – 10x + 5y^{2} + 5 –10y – 12x^{2} – 3y^{2} – 3 – 12xy + 6y + 12x = 0

– 7x^{2} + 2y^{2} + 2x – 4y – 12xy+ 7 = 0

7x^{2} + 12xy – 2y^{2} – 2x + 4y–7 = 0

∴The equation of hyperbola is7x^{2} + 12xy – 2y^{2} –2x + 4y– 7 = 0

(iv) focus is (2, -1), directrix is 2x + 3y = 1 and eccentricity= 2

Given:

Focus = (2, -1)

Directrix => 2x + 3y = 1

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

So,

^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

13{x^{2} + 4 – 4x + y^{2} + 1 +2y} = 4{4x^{2} + 9y^{2} + 1 + 12xy – 6y – 4x}

13x^{2} + 52 – 52x + 13y^{2} + 13+ 26y = 16x^{2} + 36y^{2} + 4 + 48xy – 24y – 16x

13x^{2} + 52 – 52x + 13y^{2} + 13+ 26y – 16x^{2} – 36y^{2} – 4 – 48xy + 24y +16x = 0

– 3x^{2} – 23y^{2} – 36x + 50y –48xy + 61 = 0

3x^{2} + 23y^{2} + 48xy + 36x –50y– 61 = 0

∴The equation of hyperbola is3x^{2} + 23y^{2} +48xy + 36x – 50y– 61 = 0

(v) focus is (a, 0), directrix is 2x + 3y = 1 and eccentricity =2

Given:

Focus = (a, 0)

Directrix => 2x + 3y = 1

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

So,

^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

45{x^{2} + a^{2} – 2ax + y^{2}}= 16{4x^{2} + y^{2} + a^{2} – 4xy – 2ay+ 4ax}

45x^{2} + 45a^{2} – 90ax + 45y^{2} =64x^{2} + 16y^{2} + 16a^{2} – 64xy –32ay + 64ax

45x^{2} + 45a^{2} – 90ax + 45y^{2} –64x^{2} – 16y^{2} – 16a^{2} + 64xy +32ay – 64ax = 0

19x^{2} – 29y^{2} + 154ax – 32ay –64xy – 29a^{2} = 0

∴The equation of hyperbola is19x^{2} – 29y^{2} +154ax – 32ay – 64xy – 29a^{2} = 0

(vi) focus is (2, 2), directrix is x + y = 9 and eccentricity = 2

Given:

Focus = (2, 2)

Directrix => x + y = 9

Eccentricity = 2

Now, let us find the equation of the hyperbola

Let ‘M’ be the point on directrix and P(x, y) be any pointof the hyperbola.

By using the formula,

e = PF/PM

So,

^{2} =a^{2} + b^{2} + 2ab &(a + b + c)^{2} =a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac]

x^{2} + 4 – 4x + y^{2} + 4 – 4y =2{x^{2} + y^{2} + 81 + 2xy – 18y – 18x}

x^{2} – 4x + y^{2} + 8 – 4y = 2x^{2} +2y^{2} + 162 + 4xy – 36y – 36x

x^{2} – 4x + y^{2} + 8 – 4y – 2x^{2} –2y^{2} – 162 – 4xy + 36y + 36x = 0

– x^{2} – y^{2} + 32x + 32y + 4xy– 154 = 0

x^{2} + 4xy + y^{2} – 32x – 32y +154 = 0

∴The equation of hyperbola isx^{2} + 4xy + y^{2} –32x – 32y + 154 = 0

Find the eccentricity, coordinatesof the foci, equations of directrices and length of the latus-rectum of thehyperbola.

(i) 9x^{2} – 16y^{2} = 144

(ii) 16x^{2} – 9y^{2} =-144

(iii) 4x^{2} – 3y^{2} =36

(iv) 3x^{2} – y^{2} =4

(v) 2x^{2} – 3y^{2} =5

**Answer
3** :

(i) 9x^{2} – 16y^{2} = 144

Given:

The equation => 9x^{2} – 16y^{2} =144

The equation can be expressed as:

The obtained equation is of the form

Where, a^{2} = 16, b^{2} = 9 i.e.,a = 4 and b = 3

Eccentricity is given by:

Foci: The coordinates of the foci are (0, ±be)

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

5x ∓ 16 = 0

The length of latus-rectum is given as:

2b^{2}/a

= 2(9)/4

= 9/2

(ii) 16x^{2} – 9y^{2} = -144

Given:

The equation => 16x^{2} – 9y^{2} =-144

The equation can be expressed as:

The obtained equation is of the form

Where, a^{2} = 9, b^{2} = 16 i.e.,a = 3 and b = 4

Eccentricity is given by:

Foci: The coordinates of the foci are (0, ±be)

(0, ±be) = (0, ±4(5/4))

= (0, ±5)

The equation of directrices is given as:

The length of latus-rectum is given as:

2a^{2}/b

= 2(9)/4

= 9/2

(iii) 4x^{2} – 3y^{2} = 36

Given:

The equation => 4x^{2} – 3y^{2} =36

The equation can be expressed as:

The obtained equation is of the form

Where, a^{2} = 9, b^{2} = 12 i.e.,a = 3 and b = √12

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = (±√21,0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b^{2}/a

= 2(12)/3

= 24/3

= 8

(iv) 3x^{2} – y^{2} = 4

Given:

The equation => 3x^{2} – y^{2} =4

The equation can be expressed as:

The obtained equation is of the form

Where, a = 2/√3 and b = 2

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = ±(2/√3)(2) = ±4/√3

(±ae, 0) = (±4/√3, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b^{2}/a

= 2(4)/[2/√3]

= 4√3

(v) 2x^{2} – 3y^{2} = 5

Given:

The equation => 2x^{2} – 3y^{2} =5

The equation can be expressed as:

The obtained equation is of the form

Where, a = √5/√2 and b = √5/√3

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = (±5/√6, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b^{2}/a

Find the axes, eccentricity,latus-rectum and the coordinates of the foci of the hyperbola 25x^{2} –36y^{2} = 225.

**Answer
4** :

Given:

The equation=> 25x^{2} – 36y^{2} =225

The equation can be expressed as:

The obtained equation is of the form

Where, a = 3 and b = 5/2

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

(±ae, 0) = ±3 (√61/6) = ± √61/2

(±ae, 0) = (± √61/2, 0)

The equation of directrices is given as:

The length of latus-rectum is given as:

2b^{2}/a

∴ Transverse axis = 6, conjugate axis = 5, e = √61/6, LR =25/6, foci = (± √61/2, 0)

Find the centre, eccentricity,foci and directions of the hyperbola

(i) 16x^{2} – 9y^{2} + 32x + 36y – 164 = 0

(ii) x^{2} – y^{2} +4x = 0

(iii) x^{2} – 3y^{2} –2x = 8

**Answer
5** :

(i) 16x^{2} – 9y^{2} + 32x + 36y – 164= 0

Given:

The equation => 16x^{2} – 9y^{2} +32x + 36y – 164 = 0

Let us find the centre, eccentricity, foci and directions ofthe hyperbola

By using the given equation

16x^{2} – 9y^{2} + 32x + 36y – 164= 0

16x^{2} + 32x + 16 – 9y^{2} + 36y– 36 – 16 + 36 – 164 = 0

16(x^{2} + 2x + 1) – 9(y^{2} – 4y+ 4) – 16 + 36 – 164 = 0

16(x^{2} + 2x + 1) – 9(y^{2} – 4y+ 4) – 144 = 0

16(x + 1)^{2} – 9(y – 2)^{2} = 144

Here, center of the hyperbola is (-1, 2)

So, let x + 1 = X and y – 2 = Y

The obtained equation is of the form

Where, a = 3 and b = 4

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

X = ±5 and Y = 0

x + 1 = ±5 and y – 2 = 0

x = ±5 – 1 and y = 2

x = 4, -6 and y = 2

So, Foci: (4, 2) (-6, 2)

Equation of directrix are:

∴ The center is (-1, 2), eccentricity (e) = 5/3, Foci = (4,2) (-6, 2), Equation of directrix = 5x – 4 = 0 and 5x + 14 = 0

(ii) x^{2} – y^{2} + 4x = 0

Given:

The equation => x^{2} – y^{2} +4x = 0

Let us find the centre, eccentricity, foci and directions ofthe hyperbola

By using the given equation

x^{2} – y^{2} + 4x = 0

x^{2} + 4x + 4 – y^{2} – 4 = 0

(x + 2)^{2} – y^{2} = 4

Here, center of the hyperbola is (2, 0)

So, let x – 2 = X

The obtained equation is of the form

Where, a = 2 and b = 2

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

X = ± 2√2and Y = 0

X + 2 = ± 2√2and Y = 0

X= ± 2√2– 2 and Y = 0

So, Foci = (± 2√2– 2, 0)

Equation of directrix are:

∴ The center is (-2, 0), eccentricity (e) = √2, Foci = (-2± 2√2, 0), Equation of directrix=

x + 2 = ±√2

(iii) x^{2} – 3y^{2} – 2x = 8

Given:

The equation => x^{2} – 3y^{2} –2x = 8

Let us find the centre, eccentricity, foci and directions ofthe hyperbola

By using the given equation

x^{2} – 3y^{2} – 2x = 8

x^{2} – 2x + 1 – 3y^{2} – 1 = 8

(x – 1)^{2} – 3y^{2} = 9

Here, center of the hyperbola is (1, 0)

So, let x – 1 = X

The obtained equation is of the form

Where, a = 3 and b = √3

Eccentricity is given by:

Foci: The coordinates of the foci are (±ae, 0)

X = ± 2√3and Y = 0

X – 1 = ± 2√3and Y = 0

X= ± 2√3+ 1 and Y = 0

So, Foci = (1 ± 2√3,0)

Equation of directrix are:

∴ The center is (1, 0), eccentricity (e) = 2√3/3, Foci =(1 ± 2√3, 0), Equationof directrix =

X = 1±9/2√3

Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:

(i) the distance between the foci = 16 and eccentricity = √2

(ii) conjugate axis is 5 and the distance between foci = 13

(iii) conjugate axis is 7 and passes through the point (3, -2)

**Answer
6** :

(i) the distance between the foci = 16 and eccentricity =√2

Given:

Distance between the foci = 16

Eccentricity = √2

Let us compare with the equation of the form

Distance between the foci is 2ae and b^{2} = a^{2}(e^{2} –1)

So,

2ae = 16

ae = 16/2

a√2 = 8

a = 8/√2

a^{2} = 64/2

= 32

We know that, b^{2} = a^{2}(e^{2} –1)

So, b^{2} = 32 [(√2)^{2} – 1]

= 32 (2 – 1)

= 32

The Equation of hyperbola is given as

x^{2} – y^{2} = 32

∴ The Equation of hyperbola is x^{2} – y^{2} =32

(ii) conjugate axis is 5 and the distance between foci = 13

Given:

Conjugate axis = 5

Distance between foci = 13

Let us compare with the equation of the form

Distance between the foci is 2ae and b^{2} = a^{2}(e^{2} –1)

Length of conjugate axis is 2b

So,

2b = 5

b = 5/2

b^{2} = 25/4

We know that, 2ae = 13

ae = 13/2

a^{2}e^{2} = 169/4

b^{2} = a^{2}(e^{2} – 1)

b^{2} = a^{2}e^{2} – a^{2}

25/4 = 169/4 – a^{2}

a^{2} = 169/4 – 25/4

= 144/4

= 36

The Equation of hyperbola is given as

∴ The Equation of hyperbola is 25x^{2} – 144y^{2} =900

(iii) conjugate axis is 7 and passes through the point (3,-2)

Given:

Conjugate axis = 7

Passes through the point (3, -2)

Conjugate axis is 2b

So,

2b = 7

b = 7/2

b^{2} = 49/4

The Equation of hyperbola is given as

Since it passes through points (3, -2)

∴ The Equation of hyperbola is 65x^{2} – 36y^{2} =441

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