Hyperbola

Given: Equation of directrix of a hyperbola is x – y + 3 = 0. Focus of hyperbola is (-1, 1) and eccentricity (e) = 3

To find: equation of the hyperbola

Let M be the point on directrix and P(x, y) be any point of the hyperbola

Formula used:

where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 2{x² + 1 + 2x + y² + 1 – 2y} = 9{x² + y²+ 9 + 6x – 6y – 2xy}

⇒ 2x² + 2 + 4x + 2y² + 2 – 4y = 9x² + 9y²+ 81 + 54x – 54y – 18xy

⇒ 2x² + 4 + 4x + 2y²– 4y – 9x² - 9y² - 81 – 54x + 54y + 18xy = 0

⇒ – 7x² - 7y² – 50x + 50y + 18xy – 77 = 0

⇒7x² + 7y² + 50x – 50y – 18xy + 77 = 0

This is the required equation of hyperbola

Given: Vertices (0, ±6),

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Coordinates of the vertices for a standard hyperbola is given by (0, ±b)

According to question:

b = 6 ⇒ b² = 36

We know,

a² = b²(e² – 1)

⇒ a² = 64

Hence, equation of hyperbola is:

Given: Equation of directrix of a hyperbola is x + y – 1 = 0. Focus of hyperbola is (0, 3) and eccentricity (e) = 2

To find: equation of the hyperbola

Let M be the point on directrix and P(x, y) be any point of the hyperbola

Formula used:

where e is an eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 2{x² + y² + 9 – 6y} = 4{x² + y² + 1 – 2x – 2y + 2xy}

⇒ 2x² + 2y² + 18 – 12y = 4x² + 4y²+ 4 – 8x – 8y + 8xy

⇒ 2x² + 2y² + 18 – 12y – 4x² – 4y² – 4 – 8x + 8y – 8xy = 0

⇒ – 2x² – 2y² – 8x – 4y – 8xy + 14 = 0

⇒ -2(x² + y² – 4x + 2y + 4xy – 7) = 0

⇒ x² + y² – 4x + 2y + 4xy – 7 = 0

This is the required equation of hyperbola

Given: Equation of directrix of a hyperbola is 3x + 4y + 8 = 0. Focus of hyperbola is (1, 1) and eccentricity (e) = 2

To find: equation of hyperbola

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 25{x² + 1 – 2x + y² + 1 – 2y} = 4{9x² + 16y²+ 64 + 24xy + 64y + 48x}

⇒ 25x² + 25 – 50x + 25y² + 25 – 50y = 36x² + 64y² + 256 + 96xy + 256y + 192x

⇒ 25x² + 25 – 50x + 25y² + 25 – 50y – 36x² – 64y² – 256 – 96xy – 256y – 192x = 0

⇒ – 11x² – 39y² – 242x – 306y – 96xy – 206 = 0

⇒11x² + 39y² + 242x + 306y + 96xy + 206 = 0

This is the required equation of hyperbola

Given: Equation of directrix of a hyperbola is 2x + y – 1 = 0. Focus of hyperbola is (1, 1) and eccentricity (e) =

To find: equation of hyperbola

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 5{x² + 1 – 2x + y² + 1 – 2y} = 3{4x² + y²+ 1 + 4xy – 2y – 4x}

⇒ 5x² + 5 – 10x + 5y² + 5 – 10y = 12x² + 3y² + 3 + 12xy – 6y – 12x

⇒ 5x² + 5 – 10x + 5y² + 5 – 10y – 12x² – 3y² – 3 – 12xy + 6y + 12x = 0

⇒ – 7x² + 2y² + 2x – 4y – 12xy + 7 = 0

⇒7x² – 2y² – 2x + 4y + 12xy – 7 = 0

This is the required equation of hyperbola.

Given: Equation of directrix of a hyperbola is 2x + 3y – 1 = 0. Focus of hyperbola is (2, -1) and eccentricity (e) = 2

To find: equation of hyperbola

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 13{x² + 4 – 4x + y² + 1 + 2y} = 4{4x² + 9y² + 1 + 12xy – 6y – 4x}

⇒ 13x² + 52 – 52x + 13y² + 13 + 26y = 16x² + 36y² + 4 + 48xy – 24y – 16x

⇒ 13x² + 52 – 52x + 13y² + 13 + 26y – 16x² – 36y² – 4 – 48xy + 24y + 16x = 0

⇒ – 3x² – 23y² – 36x + 50y – 48xy + 61 = 0

⇒3x² + 23y² + 36x – 50y + 48xy – 61 = 0

This is the required equation of hyperbola.

Given: Equation of directrix of a hyperbola is 2x – y + a = 0. Focus of hyperbola is (a, 0) and eccentricity (e) =

To find: equation of hyperbola

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ 117{x² + a² – 2ax + y²} = 16{4x² + y² + a² – 4xy – 2ay + 4ax}

⇒ 117x² + 117a² – 234ax + 117y² = 64x² + 16y² + 16a² – 64xy – 32ay + 64ax

⇒ 117x² + 117a² – 234ax + 117y² – 64x² – 16y² – 16a² + 64xy + 32ay – 64ax = 0

⇒ 53x² + 101y² – 298ax + 32ay + 64xy + 111a² = 0

This is the required equation of hyperbola.

Given: Equation of directrix of a hyperbola is x + y – 9 = 0. Focus of hyperbola is (2, 2) and eccentricity (e) = 2

To find: equation of hyperbola

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab &

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac}

⇒ x² + 4 – 4x + y² + 4 – 4y = 2{x² + y² + 81 + 2xy – 18y – 18x}

⇒ x² – 4x + y² + 8 – 4y = 2x² + 2y² + 162 + 4xy – 36y – 36x

⇒ x² – 4x + y² + 8 – 4y – 2x² – 2y² – 162 – 4xy + 36y + 36x = 0

⇒ – x² – y² + 32x + 32y + 4xy – 154 = 0

⇒x² + y² – 32x – 32y + 4xy + 154 = 0

This is the required equation of hyperbola.

Given: 9x² – 16y² = 144

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Equation of directrix are:

Length of latus rectum is

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Foci: (±5, 0)

Equation of directrix are:

Length of latus rectum,

Given: 16x² – 9y² = -144

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (0, ±be)

The equation of directrix are:

Length of latus rectum is

Here, a = 3 and b = 4

⇒ c = 5

Therefore,

Foci: (0, ±5)

The equation of directrix are:

Length of latus rectum,

Given: 4x² – 3y² = 36

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = 3 and b =

Therefore,

The equation of directrix are:

Length of latus rectum,

= 8

Given: 3x² – y² = 4

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = and b = 2

Therefore,

The equation of directrix are:

Length of latus rectum,

Given: 2x² – 3y² = 5

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = and b =

Therefore,

The equation of directrix are:

Length of latus rectum,

Given: 2x² – 3y² = 5

To find: eccentricity(e), coordinates of the foci f(m,n), equation of directrix, length of latus-rectum of hyperbola.

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = 3 and b =

Therefore,

The equation of directrix are:

Length of latus rectum,

Given: 16x² – 9y² + 32x + 36y – 164 = 0

To find: center, eccentricity(e), coordinates of the foci f(m,n), equation of directrix.

16x² – 9y² + 32x + 36y – 164 = 0

⇒ 16x² + 32x + 16 – 9y² + 36y – 36 – 16 + 36 – 164 = 0

⇒ 16(x² + 2x + 1) – 9(y² – 4y + 4) – 16 + 36 – 164 = 0

⇒ 16(x² + 2x + 1) – 9(y² – 4y + 4) – 144 = 0

⇒ 16(x + 1)² – 9(y – 2)² = 144

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

Let x + 1 = X and y – 2 = Y

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = 3 and b = 4

Therefore,

⇒ X = ±5 and Y = 0

⇒ x + 1 = ±5 and y – 2 = 0

⇒ x = ±5 – 1 and y = 2

So, Foci: (±5 – 1, 2)

Equation of directrix are:

Given: x² – y² + 4x = 0

To find: center, eccentricity(e), coordinates of the foci f(m,n), equation of directrix.

x² – y² + 4x = 0

⇒ x² + 4x + 4 – y² – 4 = 0

⇒ (x + 2)² – y² = 4

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

Let x – 2 = X

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci are given by (±ae, 0)

The equation of directrix are

Length of latus rectum is

Here, a = 2 and b = 2

Therefore,

⇒ and y = 0

⇒ and y = 0

⇒ and y = 0

Equation of directrix are:

Given: x² – 3y² – 2x = 8

To find: center, eccentricity(e), coordinates of the foci f(m,n), equation of directrix.

x² – 3y² – 2x = 8

⇒ x² – 2x + 1 – 3y² – 1 = 8

⇒ (x – 1)² – 3y² = 9

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

Let x – 1 = X

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Equation of directrix are:

Length of latus rectum is

Here, a = 3 and b =

Therefore,

⇒ and y = 0

⇒ and y = 0

⇒ and y = 0

Equation of directrix are:

Given: the distance between the foci = 16 and eccentricity

To find: the equation of the hyperbola

Formula used:

For hyperbola:

Distance between the foci is 2ae and b² = a²(e² – 1)

Therefore

2ae = 16

b² = a²(e² – 1)

Equation of hyperbola is:

Hence, required equation of hyperbola is x² – y² = 32

Given: the distance between the foci = 13 and conjugate axis is 5

To find: the equation of the hyperbola

Formula used:

For hyperbola:

Distance between the foci is 2ae and b² = a²(e² – 1)

Length of conjugate axis is 2b

Therefore

2ae = 13

b² = a²(e² – 1)

⇒ b² = a²e² – a²

Equation of hyperbola is:

Hence, required equation of hyperbola is 25x² – 144y² = 900

Given: conjugate axis is 5 and passes through the point (3, -2)

To find: the equation of the hyperbola

Formula used:

For hyperbola:

Conjugate axis is 2b

Therefore

The equation of hyperbola is:

Since it passes through (3, -2)

The equation of hyperbola:

Hence, required equation of hyperbola is 65x² – 36y² = 441

Given: Foci are (6, 4) and (-4, 4) and eccentricity is 2

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two foci.

Distance between the foci is 2ae and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having foci (6, 4) and (-4, 4) is given by

= (1, 4)

The distance between the foci is 2ae, and Foci are (6, 4) and (-4, 4)

{∵ e = 2}

b² = a²(e² – 1)

The equation of hyperbola:

⇒ 12(x² + 1 – 2x) – 4(y² + 16 – 8y) = 75

⇒ 12x² + 12 – 24x – 4y² – 64 + 32y – 75 = 0

⇒ 12x² – 4y² – 24x + 32y – 127 = 0

Hence, required equation of hyperbola is 12x² – 4y² – 24x + 32y – 127 = 0

Given: Vertices are (-8, -1) and (16, -1) and focus is (17, -1)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two vertices

The distance between two vertices is 2a

The distance between the foci and vertex is ae – a and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having vertices (-8, -1) and (16, -1) is given by

= (4, -1)

The distance between two vertices is 2a and vertices are (-8, -1) and (16, -1)

The distance between the foci and vertex is ae – a, Foci is (17, -1) and the vertex is (16, -1)

b² = a²(e² – 1)

The equation of hyperbola:

⇒ 25(x² + 16 – 8x) – 144(y² + 1 + 2y) = 3600

⇒ 25x² + 400 – 200x – 144y² – 144 – 288y – 3600 = 0

⇒ 25x² – 144y² – 200x – 288y – 3344 = 0

Hence, required equation of hyperbola is 25x² – 144y² – 200x – 288y – 3344 = 0

Given: Foci are (4, 2) and (8, 2) and eccentricity is 2

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two foci.

Distance between the foci is 2ae and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having foci (4, 2) and (8, 2) is given by

= (6, 2)

The distance between the foci is 2ae and Foci are (4, 2) and (8, 2)

{∵ e = 2}

b² = a²(e² – 1)

The equation of hyperbola:

⇒ 3(x² + 36 – 12x) – (y² + 4 – 4y) = 3

⇒ 3x² + 108 – 36x – y² – 4 + 4y – 3 = 0

⇒ 3x² – y² – 36x + 4y + 101 = 0

Hence, required equation of hyperbola is 3x² – y² – 36x + 4y + 101 = 0

Given: Vertices are (0, ± 7) and foci are

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Vertices of the hyperbola are given by (0, ±b)

Foci of the hyperbola are given by (0, ±be)

Vertices are (0, ±7) and foci are

Therefore,

a² = b²(e² – 1)

The equation of hyperbola:

⇒ 9x² – 7y² = -343

⇒ 9x² – 7y² + 343 = 0

Hence, required equation of hyperbola is 9x² – 7y² + 343 = 0

Given: Vertices are (± 6, 0) and one of the directrices is x = 4

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Vertices of the hyperbola are given by (±a, 0)

The equation of the directrices:

Vertices are (± 6, 0) and one of the directrices is x = 4

Therefore,

b² = a²(e² – 1)

⇒ b² = 45

The equation of hyperbola:

⇒ 5x² – 4y² = 180

⇒ 5x² – 4y² – 180 = 0

Hence, required equation of hyperbola is 5x² – 4y² – 180 = 0

Given: Foci are (2, 0) and (-2, 0) and eccentricity is

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two foci.

Distance between the foci is 2ae and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having Foci (2, 0) and (-2, 0) is given by

= (0, 0)

The distance between the foci is 2ae, and Foci are (2, 0) and (-2, 0)

b² = a²(e² – 1)

The equation of hyperbola:

⇒ 45x² – 36y² – 80 = 0

Hence, required equation of hyperbola is 45x² – 36y² – 80 = 0

Given: the length of whose conjugate axis is of the length of the transverse axis

To find: eccentricity of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Length of the conjugate axis is 2b and length of transverse axis is 2a

According to question:

We know,

Hence, the eccentricity of the hyperbola is

Given: Vertices are (4, 2) and (2, 2), the focus is (5, 2) and centre (3, 2)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two vertices

The distance between two vertices is 2a

The distance between the foci and vertex is ae – a and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

The distance between two vertices is 2a and vertices are (4, 2) and (2, 2)

The distance between the foci and vertex is ae – a, Foci is (5, 2) and the vertex is (4, 2)

⇒ 1 = e – 1

⇒ e = 1 + 1

⇒ e = 2

b² = a²(e² – 1)

The equation of hyperbola:

{∵ Centre (3, 2)}

⇒ 3(x² + 9 – 6x) – (y² + 4 – 4y) = 3

⇒ 3x² + 27 – 18x – y² – 4 + 4y – 3 = 0

⇒ 3x² – y² – 18x + 4y + 20 = 0

Hence, required equation of hyperbola is 3x² – y² – 18x + 4y + 20 = 0

Given: Foci is (4, 2), e = 2 and center at (6, 2)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Center is the mid-point of two vertices

The distance between two vertices is 2a

The distance between the foci and vertex is ae – a and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Therefore

Let one of the two foci is (m, n) and the other one is (4, 2)

Since, Centre(6, 2)

Foci are (4, 2) and (8, 2)

The distance between the foci is 2ae and Foci are (4, 2) and (8, 2)

⇒ a² = 1

b² = a²(e² – 1)

⇒ b² = 4 – 1

⇒ b² = 3

The equation of hyperbola:

⇒ 3(x² + 36 – 12x) – (y² + 4 – 4y) = 3

⇒ 3x² + 108 – 36x – y² – 4 + 4y – 3 = 0

⇒ 3x² – y² – 36x + 4y + 101 = 0

Hence, required equation of hyperbola is 3x² – y² – 36x + 4y + 101 = 0

Given: Axis of the hyperbola are equal, i.e. a = b

To prove: SP.S’P = CP²

Formula used:

The standard form of the equation of the hyperbola is,

Foci of the hyperbola are given by (±ae, 0)

Let P (m, n) be any point on the hyperbola

The distancebetween two points (m, n) and (a, b) is given by

C is Centre with coordinates (0, 0)

Now,

{∵ a² = m² – n²}

From (i):

Taking square root both sides:

Hence Proved

Given: Vertices are (±2, 0) and foci are (±3, 0)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Vertices of the hyperbola are given by (±a, 0)

Foci of the hyperbola are given by (±ae, 0)

Vertices are (±2, 0) and foci are (±3, 0)

Therefore,

a = 2 and ae = 3

⇒ 2 × e = 3

b² = a²(e² – 1)

⇒ b² = 5

The equation of hyperbola:

⇒ 5x² – 4y² = 20

⇒ 5x² – 4y² – 20 = 0

Hence, required equation of hyperbola is 5x² – 4y² – 20 = 0

Given: Vertices are (0, ±5) and foci are (0, ±8)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Vertices of the hyperbola are given by (0, ±b)

Foci of the hyperbola are given by (0, ±be)

Vertices are (0, ±5) and foci are (0, ±8)

Therefore,

a² = b²(e² – 1)

⇒ a² = 39

The equation of hyperbola:

Hence, the required equation of the hyperbola is

Given: Vertices are (0, ±3) and foci are (0, ±5)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Vertices of the hyperbola are given by (0, ±b)

Foci of the hyperbola are given by (0, ±be)

Vertices are (0, ±3) and foci are (0, ±5)

Therefore,

a² = b²(e² – 1)

⇒ a² = 16

The equation of hyperbola:

Hence, the required equation of the hyperbola is

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Length of transverse axis is 2a

Coordinates of the foci for a standard hyperbola is given by (±ae, 0)

According to question:

2a = 8 and ae = 5

2a = 8

⇒ a = 4

⇒ a² = 16

ae = 5

We know,

b² = a²(e² – 1)

⇒ b² = 4(21)

⇒ b² = 84

Hence, the equation of the hyperbola is:

Given: foci (0, ±13) and the conjugate axis is 24

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Length of the conjugate axis is 2b

Coordinates of the foci for a standard hyperbola is given by (0, ±be)

According to question:

2b = 24 and be = 13

2b = 24

⇒ b = 12

⇒ b² = 144

be = 12

We know,

a² = b²(e² – 1)

⇒ b² = 25

Hence, the equation of the hyperbola is:

Given: Foci and the latus-rectum = 8

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Coordinates of the foci for a standard hyperbola is given by (±ae, 0)

Length of latus rectum is

According to the question:

We know,

b² = a²(e² – 1)

⇒ 4a = 45 – a²

⇒ a² + 4a – 45 = 0

⇒ a² + 9a – 5a – 45 = 0

⇒ a(a + 9) – 5(a + 9) = 0

⇒ (a + 9)(a – 5) = 0

⇒ a = -9 or a = 5

Since a is a distance, and it can’t be negative

⇒ a = 5

⇒ a² = 25

b² = 4a

⇒ b² = 4(5)

⇒ b² = 20

Hence, equation of hyperbola is:

Given: Foci (±4, 0), the latus-rectum = 12

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Coordinates of the foci for a standard hyperbola is given by (±ae, 0)

Length of latus rectum is

According to the question:

ae = 4

We know,

b² = a²(e² – 1)

⇒ 6a = 16 – a²

⇒ a² + 6a – 16 = 0

⇒ a² + 8a – 2a – 16 = 0

⇒ a(a + 8) – 2(a + 8) = 0

⇒ (a + 8)(a – 2) = 0

⇒ a = -8 or a = 2

Since a is a distance, and it can’t be negative,

⇒ a = 2

⇒ a² = 4

b² = 6a

⇒ b² = 6(2)

⇒ b² = 12

Hence, equation of hyperbola is:

Given: passing through (2, 3)

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Coordinates of the foci for a standard hyperbola is given by (0, ±be)

According to the question:

⇒ b²e² = 10

Since (2, 3) passing through hyperbola

Therefore,

{∵ a² = b²(e² – 1)}

⇒ 90 – 13b² = (10 – b²)b²

⇒ 90 – 13b² = 10b² – b⁴

⇒ 90 – 13b² – 10b² + b⁴ = 0

⇒ b⁴ – 23b² + 90 = 0

⇒ b⁴ – 18b² – 5b² + 90 = 0

⇒ b²(b² – 18) – 5(b² – 18) = 0

⇒ (b² – 18)(b² – 5) = 0

⇒ b² = 18 or 5

Case 1: b² = 18 and b²e² = 10

a² = b²(e² – 1)

⇒ a² = b²e² – b²

⇒ a² = 10 – 18

⇒ a² = – 8

Hence, equation of hyperbola is:

Case 2: b² = 5 and b²e² = 10

a² = b²(e² – 1)

⇒ a² = b²e² – b²

⇒ a² = 10 – 5

⇒ a² = 5

Hence, equation of hyperbola is:

Given: Foci (0, ±12), the latus-rectum = 36

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

Coordinates of the foci for a standard hyperbola is given by (0, ±be)

Length of latus rectum is

According to the question:

be = 12

We know,

a² = b²(e² – 1)

⇒ 18b = 144 – b²

⇒ b² + 18b – 144 = 0

⇒ b² + 24b – 6b – 144 = 0

⇒ b(b + 24) – 6(b + 24) = 0

⇒ (b + 24)(b – 6) = 0

⇒ b = -24 or b = 6

Since b is a distance, and it can’t be negative

⇒ b = 6

⇒ b² = 36

a² = 18b

⇒ a² = 18(6)

⇒ b² = 108

Hence, equation of hyperbola is:

Given: Distance between foci is 16 and eccentricity is

To find: equation of the hyperbola

Formula used:

The standard form of the equation of the hyperbola is,

The distance between foci is given by 2ae

According to question:

2ae = 16

⇒ a² = 32

We know,

b² = a²(e² – 1)

⇒ b² = 32

Hence, the equation of the hyperbola is:

⇒ x² – y² = 32

To prove: the set of all points under given conditions represents a hyperbola

Let a point P be (x, y) such that the difference of their distances from (4, 0) and (-4, 0) is always equal to 2.

Formula used:

The distance between two points (m, n) and (a, b) is given by

The distance of P(x, y) from (4, 0) is

The distance of P(x, y) from (-4, 0) is

Since, the difference of their distances from (4, 0) and (-4, 0) is always equal to 2

Therefore,

Squaring both sides:

Squaring both sides:

⇒ 16x² + 1 – 8x = (x – 4)² + y²

⇒ 16x² + 1 – 8x = x² + 16 – 8x + y²

⇒ 16x² + 1 – 8x – x² – 16 + 8x – y² = 0

⇒ 15x² – y² – 15 = 0

Hence, required equation of hyperbola is 15x² – y² – 15 = 0

Given: 9x² – 16y² = 144

To find: eccentricity(e)

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Hence, eccentricity is

Given: Latus-rectum is half of its transverse axis

To find: eccentricity of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Length of transverse axis is 2a

Latus-rectum of the hyperbola is

According to question:

Latus-rectum is half of its transverse axis

We know,

Hence, eccentricity is

Given: 9x² – 16y² = 144

To find: coordinates of the foci f(m,n)

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Foci: (±5, 0)

Given: Distance between foci is 16 and eccentricity is

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Distance between foci is given by 2ae

According to question:

2ae = 16

⇒ a² = 32

We know,

b² = a²(e² – 1)

⇒ b² = 32

Hence, equation of hyperbola is:

⇒ x² – y² = 32

To find: value of b²

Given: foci of given ellipse and hyperbola coincide

Formula used:

Coordinates of the foci for standard ellipse is given by (±c₁, 0) where

Coordinates of the foci for standard hyperbola is given by (±c₂, 0) where

Since, their foci coincide

Here a₁ = 4, b₁ = b,

⇒ b² = 16 – 9

⇒ b² = 7

Given: 16x² – 9y² = 144

To find: length of latus-rectum of hyperbola.

16x² – 9y² = 144

Formula used:

For hyperbola

Length of latus rectum is

Here, a = 3 and b = 4

Length of latus rectum,

Given: latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex

To find: eccentricity of hyperbola

Let B is vertex of hyperbola and A and C are point of intersection of latus-rectum and hyperbola

Standard equation of hyperbola is

Since, A and C lie on hyperbola

Therefore

As angle between AB and AC is 90°

⇒ Slope_AB × Slope_BC = -1

From (i):

{∵ b² = a²(e² – 1)}

⇒ (e² – 1)² = (e + 1)²

⇒ e⁴ + 1 – 2e² = e² + 1 + 2e

⇒ e⁴ + 1 – 2e² – e² – 1 – 2e = 0

⇒ e⁴ – 3e² – 2e = 0

⇒ e(e³ – 3e – 2) = 0

⇒ e(e– 2)(e + 1)² = 0

⇒ e = 0 or 2 or -1

But e should be greater than or equal to 1 for hyperbola

⇒ e = 2

Hence, eccentricity of hyperbola is 2

Given: Hyperbola is x = 8 sec θ and y = 8 tan θ

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Distance between directrix is given by

x = 8 sec θ and y = 8 tan θ

We know,

sec² θ – tan² θ = 1

Here a = 8 and b = 8

Now,

Hence, distance between directrix,

Given: Vertices are (±3, 0) and foci are (±5, 0)

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Vertices of hyperbola are given by (±a, 0)

Foci of hyperbola are given by (±ae, 0)

Vertices are (±3, 0) and foci are (±5, 0)

Therefore,

a = 3 and ae = 5

⇒ 3 × e = 5

b² = a²(e² – 1)

⇒ b² = 16

Equation of hyperbola:

Hence, required equation of hyperbola is

Given: e₁ and e₂ are respectively the eccentricities of the ellipse and the hyperbola

To find: value of 2e₁² + e₂²

Eccentricity(e) of hyperbola is given by,

Here a = 3 and b = 2

Therefore,

For ellipse:

Eccentricity(e) of ellipse is given by,

Therefore,

Substituting values from (1) and (2) in 2e₁² + e₂²

2e₁² + e₂²

= 3

Hence, value of 2e₁² + e₂² is 3

Given: Vertices are (±3, 0) and foci are (±5, 0)

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Vertices of hyperbola are given by (±a, 0)

Foci of hyperbola are given by (±ae, 0)

Vertices are (±3, 0) and foci are (±5, 0)

Therefore,

a = 3 and ae = 5

⇒ 3 × e = 5

b² = a²(e² – 1)

⇒ b² = 16

Equation of hyperbola:

⇒ 16x² – 9y² = 144

Hence, required equation of hyperbola is 16x² – 9y² = 144

Given: e₁ and e₂ are respectively the eccentricities of the ellipse and the hyperbola

To find: value of 2e₁² + e₂²

Eccentricity(e) of hyperbola is given by,

Here a = 3 and b = 2

Therefore,

For ellipse:

Eccentricity(e) of ellipse is given by,

Therefore,

Substituting values from (1) and (2) in 2e₁² + e₂²

2e₁² + e₂²

= 3

Hence, value of 2e₁² + e₂² is 3

Given: Hyperbola is x = 8 sec θ and y = 8 tan θ

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Distance between directrix is given by

x = 8 sec θ and y = 8 tan θ

We know,

sec² θ – tan² θ = 1

Here a = 8 and b = 8

Now,

Hence, distance between directrix,

Given: Equation of directrix of a hyperbola is x – y + 1= 0. Focus of hyperbola is (1, -1) and eccentricity (e) is

To find: equation of conic

Let M be the point on directrix and P(x, y) be any point of hyperbola

Formula used:

where e is eccentricity, PM is perpendicular from any point P on hyperbola to the directrix

Therefore,

Squaring both sides:

{∵ (a – b)² = a² + b² + 2ab}

⇒ x² + 1 – 2x + y² + 1 + 2y = x² + y² + 1 – 2xy + 2x – 2y

⇒ x² + 1 – 2x + y² + 1 + 2y – x² – y² + 2xy – 1 – 2x + 2y = 0

⇒ 2xy – 4x + 4y + 1 = 0

This is the required equation of hyperbola

Given: 9x² – 16y² = 144

To find: eccentricity(e)

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Hence, eccentricity is

Given: 16x² – 9y² = 144

To find: length of latus-rectum of hyperbola.

16x² – 9y² = 144

Formula used:

For hyperbola

Length of latus rectum is

Here, a = 3 and b = 4

Length of latus rectum,

We know it by the fact that when difference in distances is constant, it forms as hyperbola

Given: Latus-rectum is half of its transverse axis

To find: eccentricity of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Length of transverse axis is 2a

Latus-rectum of the hyperbola is

According to question:

Latus-rectum is half of its transverse axis

We know,

Hence, eccentricity is

Given: x² – 4y² = 1

To find: eccentricity(e)

x² – 4y² = 1

Formula used:

For hyperbola

Eccentricity(e) is given by,

Here, a = 1 and

Therefore,

Hence, eccentricity is

This is definition of eccentricity.

Eccentricity is difference of the focal distances of any point on the hyperbola

Given: 9x² – 16y² = 144

To find: coordinates of the foci f(m,n)

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Foci: (±5, 0)

Given: Distance between foci is 16 and eccentricity is

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Distance between foci is given by 2ae

According to question:

2ae = 16

⇒ a² = 32

We know,

b² = a²(e² – 1)

⇒ b² = 32

Hence, equation of hyperbola is:

⇒ x² – y² = 32

Given: e₁ and e₂ are respectively the eccentricities of 9x² + 4y² = 36 and 9x² – 4y² = 36 respectively

To find: e₁² – e₂²

9x² – 4y² = 36

Eccentricity(e) of hyperbola is given by,

Here a = 2 and b = 3

Therefore,

For ellipse:

9x² + 4y² = 36

Eccentricity(e) of ellipse is given by,

Here a = 2 and b = 3

Therefore,

Substituting values from (1) and (2) in 2e₁² + e₂²

e₁² – e₂²

⇒ e₂² – e₂²

Hence, value of 2 < e₂² – e₁² < 3

Given: e₁ and e₂ are respectively the eccentricities of x² – y² sec² α = 5 and x² sec² α + y² = 25 respectively

To find: value of α

x² – y² sec² α = 5

Eccentricity(e) of hyperbola is given by,

Therefore,

For ellipse:

x² sec² α + y² = 25

Eccentricity(e) of ellipse is given by,

Therefore,

According to question:

Eccentricity of given hyperbola is time eccentricity of given ellipse

From (1) and (2):

Squaring both sides:

⇒ 1 + cos² α = 3(1 – cos² α)

⇒ 1 + cos² α = 3 – 3 cos² α

⇒ 3 cos² α + cos² α = 3 – 1

⇒ 4 cos² α = 2

Given: Foci are (6, 4) and (-4, 4) and eccentricity is 2

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Center is the mid-point of two foci.

Distance between the foci is 2ae and b² = a²(e² – 1)

Distance between two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Center of hyperbola having foci (6, 4) and (-4, 4) is given by

= (1, 4)

Distance between the foci is 2ae and Foci are (6, 4) and (-4, 4)

{∵ e = 2}

b² = a²(e² – 1)

Equation of hyperbola:

Hence, required equation of hyperbola is

Given: A straight line x – 3y = 1 intercepts hyperbola x² – 4y² = 1

To find: Length of the intercepted line

Formula used:

Distance between two points (m, n) and (a, b) is given by

Firstly we will find point of intersections of given line and hyperbola

x – 3y = 1 ⇒ x = 1 + 3y

x² – 4y² = 1

⇒ (1 + 3y)² – 4y² = 1

⇒ 1 + 9y² + 6y – 4y² = 1

⇒ 5y² + 6y = 0

⇒ y(5y + 6) = 0

⇒ y = 0 or 5y + 6 = 0

Now, x = 1 + 3y

So, Point of intersections are A(1, 0) and

Distance between point of intersections is

Given: 2x² – 3y² = 5

To find: coordinates of the foci f(m,n)

2x² – 3y² = 5

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Therefore,

Given: Equation of hyperbola

_{To find:Eccentricity of the hyperbola}

Squaring both sides:

Squaring both sides:

_{From (1) and (2):}

Formula used:

For hyperbola

Eccentricity(e) is given by,

Here a = a, b = a

Therefore,

_{Hence, the}eccentricity of the hyperbola is

Given: Foci is (4, 2), e = 2 and center at (6, 2)

To find: equation of the hyperbola

Formula used:

Standard form of the equation of hyperbola is,

Center is the mid-point of two vertices

The distance between two vertices is 2a

The distance between the foci and vertex is ae – a and b² = a²(e² – 1)

The distancebetween two points (m, n) and (a, b) is given by

Mid-point theorem:

Mid-point of two points (m, n) and (a, b) is given by

Therefore

Let one of the two foci is (m, n) and the other one is (4, 2)

Since, Centre(6, 2)

Foci are (4, 2) and (8, 2)

The distance between the foci is 2ae and Foci are (4, 2) and (8, 2)

⇒ a² = 1

b² = a²(e² – 1)

⇒ b² = 4 – 1

⇒ b² = 3

Equation of hyperbola:

⇒ 3(x – 6)² – (y – 2)² = 3

Hence, required equation of hyperbola is 3(x – 6)² – (y – 2)² = 3

Given: A hyperbola is formed by the locus of the point of intersection of lines

To find: Eccentricity of the hyperbola

Multiply by λ:

Adding (1) and (2):

Now, From (1):

Squaring both sides:

From (3) and (4):

Formula used:

For hyperbola

Eccentricity(e) is given by,

⇒ c = 8

Therefore,

⇒ e = 2

Hence, eccentricity of hyperbola is 2