Ellipse

Given that we need to find the equation of the ellipse whose focus is S(1, - 2) and directrix(M) is 3x - 2y + 5 = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on the ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 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.

Given that we need to find the equation of the ellipse whose focus is S(0,1) and directrix(M) is x + y = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on the ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 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.

Given that we need to find the equation of the ellipse whose focus is S(- 1,1) and directrix(M) is x - y + 3 = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on the ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 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.

Given that we need to find the equation of the ellipse whose focus is S(- 2,3) and directrix(M) is 2x + 3y + 4 = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on the ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 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.

Given that we need to find the equation of the ellipse whose focus is S(1, 2) and directrix(M) is 3x + 4y - 5 = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on the ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 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.

Given the equation of the ellipse is 4x² + 9y² = 1.

We need to find the eccentricity, coordinates of foci and length of latus rectum.

Given equation can be rewritten as .

We know for the ellipse (a²>b²)

⇒

⇒ Coordinates of foci (±ae,0)

⇒ Length of latus rectum =

Here and , a²>b²

⇒

⇒

⇒

⇒

⇒

⇒

⇒ Length of latus rectum (L) =

⇒

∴ The eccentricity is , foci are and length of the latus rectum is .

Given the equation of the ellipse is 5x² + 4y² = 1.

We need to find the eccentricity, coordinates of foci and length of latus rectum.

Given equation can be rewritten as .

We know for the ellipse (b²>a²)

⇒

⇒ Coordinates of foci (0,±be)

⇒ Length of latus rectum =

Here and , b²>a²

⇒

⇒

⇒

⇒

⇒

⇒ Length of latus rectum (L) =

⇒

∴ The eccentricity is , foci are and length of the latus rectum is .

Given the equation of the ellipse is 4x² + 3y² = 1.

We need to find the eccentricity, coordinates of foci and length of latus rectum.

Given equation can be rewritten as .

We know for the ellipse (b²>a²)

⇒

⇒ Coordinates of foci (0,±be)

⇒ Length of latus rectum =

Here and , b²>a²

⇒

⇒

⇒

⇒

⇒

⇒

⇒ Length of latus rectum (L) =

⇒

∴ The eccentricity is , foci are and length of the latus rectum is .

Given the equation of the ellipse is 25x² + 16y² = 1600.

We need to find the eccentricity, coordinates of foci and length of latus rectum.

Given equation can be rewritten as

⇒ .

We know for the ellipse (b²>a²)

⇒

⇒ Coordinates of foci (0,±be)

⇒ Length of latus rectum =

Here a² = 64 and b² = 100, b²>a²

⇒

⇒

⇒

⇒

⇒

⇒ foci = (0,±6)

⇒ Length of latus rectum (L) =

⇒

∴ The eccentricity is , foci are (0,±6) and length of the latus rectum is .

Given the equation of the ellipse is 9x² + 25y² = 225.

We need to find the eccentricity, coordinates of foci and length of latus rectum.

Given equation can be rewritten as

⇒ .

We know for the ellipse (a²>b²)

⇒

⇒ Coordinates of foci (±ae,0)

⇒ Length of latus rectum =

Here a² = 25 and b² = 9, a²>b²

⇒

⇒

⇒

⇒

⇒

⇒ Length of latus rectum (L) =

⇒

∴ The eccentricity is , foci are and length of the latus rectum is .

Given that we need to find the equation of the ellipse (whose axes are x = 0 and y = 0) which passes through the point (- 3,1) and has eccentricity .

We know that the equation of the ellipse whose axes are x and y - axis is . ..... - - - - - (1)

Let us assume a²>b².

We know that eccentricity(e) =

⇒

⇒

⇒

⇒ ..... .... (2)

Substituting (2) in (1) we get,

⇒

⇒

⇒ 3x² + 5y² = 3a²

This curve passes through the point (- 3,1). Substituting in the curve we get,

⇒ 3(- 3)² + 5(1)² = 3a²

⇒ 3(9) + 5 = 3a²

⇒ 32 = 3a²

⇒

⇒

⇒

The equation of the ellipse is:

⇒

⇒

⇒ 3x² + 5y² = 32

∴ The equation of the ellipse is 3x² + 5y² = 32.

Given that we need to find the equation of the ellipse whose eccentricity is and foci (±2,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

We know that foci = (±ae,0)

⇒ ae = 2

⇒

⇒ a = 4

⇒ a² = 16

⇒

⇒ b² = 12

The equation of the ellipse is

⇒

⇒

⇒ 3x² + 4y² = 48

∴ The equation of the ellipse is 3x² + 4y² = 48.

Given that we need to find the equation of the ellipse whose eccentricity is and length of latus rectum is 5.

Let us assume the equation of the ellipse as (a²>b²).

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

We know that the length of the latus rectum is .

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒

⇒ 20x² + 36y² = 405

∴ The equation of the ellipse is 20x² + 36y² = 405.

Given that we need to find the equation of the ellipse whose eccentricity is and the semi - major axis is 4.

Let us assume the equation of the ellipse as (a²>b²).

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

We know that the length of the semi - major axis is a

⇒ a = 4

⇒ a² = 16

⇒

⇒ b² = 12

The equation of the ellipse is

⇒

⇒

⇒ 3x² + 4y² = 48

∴ The equation of the ellipse is 3x² + 4y² = 48.

Given that we need to find the equation of the ellipse whose eccentricity is and the major axis is 12.

Let us assume the equation of the ellipse as (a²>b²).

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

We know that length of ,ajor axis is 2a.

⇒ 2a = 12

⇒ a = 6

⇒ a² = 36

⇒

⇒ b² = 27

The equation of the ellipse is

⇒

⇒

⇒ 3x² + 4y² = 108

∴ The equation of the ellipse is 3x² + 4y² = 108.

Given that we need to find the equation of the ellipse passing through the points (1,4) and (- 6,1).

Let us assume the equation of the ellipse as (a²>b²). ..... .... (1)

Substituting the point (1,4) in (1) we get

⇒

⇒

⇒

⇒ b² + 16a² = a² b² ..... - - (2)

Substituting the point (- 6,1) in (1) we get

⇒

⇒

⇒

⇒ a² + 36b² = a²b² ..... - - (3)

(3)×16 - (2)

⇒ (16a² + 576b²) - (b² + 16a²) = (16a²b² - a²b²)

⇒ 575b² = 15a²b²

⇒ 15a² = 575

⇒

From (2)

⇒

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒ 3x² + 7y² = 115

∴ The equation of the ellipse is 3x² + 7y² = 115.

Given that we need to find the equation of the ellipse whose vertices are (±5,0) and foci (±4,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that vertices of the ellipse are (±a,0)

⇒ a = 5

⇒ a² = 25

We know that foci = (±ae,0)

⇒ ae = 4

⇒ 5e = 4

⇒

We know that eccentricity

⇒

⇒

⇒

⇒ b² = 9

The equation of the ellipse is

⇒

⇒

⇒ 9x² + 25y² = 225

∴ The equation of the ellipse is 9x² + 25y² = 225.

Given that we need to find the equation of the ellipse whose vertices are (0,±13) and foci (±4,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that vertices of the ellipse are (0,±b)

⇒ b = 13

⇒ b² = 169

We know that foci = (±ae,0)

⇒ ae = 4

We know that eccentricity

⇒

⇒ 16 = a² - 169

⇒ a² = 185

The equation of the ellipse is

⇒

∴ The equation of the ellipse is .

Given that we need to find the equation of the ellipse whose vertices are (±6,0) and foci (±4,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that vertices of the ellipse are (±a,0)

⇒ a = 6

⇒ a² = 36

We know that foci = (±ae,0)

⇒ ae = 4

⇒ 6e = 4

⇒

We know that eccentricity

⇒

⇒

⇒

⇒ b² = 20

The equation of the ellipse is

⇒

⇒

⇒ 5x² + 9y² = 180

∴ The equation of the ellipse is 5x² + 9y² = 180.

Given that we need to find the equation of the ellipse whose ends of the major axis is (±3,0) and ends of the minor axis is (0,±2).

Let us assume the equation of the ellipse as (a²>b²).

We know that ends of the major axis of the ellipse are (±a,0)

⇒ a = 3

⇒ a² = 9

We know that ends of minor axis of the ellipse are (0,±b)

⇒ b = 2

⇒ b² = 4

The equation of the ellipse is

⇒

⇒

⇒ 4x² + 9y² = 36

∴ The equation of the ellipse is 4x² + 9y² = 36.

Given that we need to find the equation of the ellipse whose ends of major axis are and ends of the minor axis are (±1,0).

Let us assume the equation of the ellipse as (b²>a²).

We know that ends of the major axis of the ellipse are (0,±b)

⇒

⇒ b² = 5

We know that ends of the minor axis of the ellipse are (±a,0)

⇒ a = 1

⇒ a² = 1

The equation of the ellipse is

⇒

⇒

⇒ 5x² + y² = 5

∴ The equation of the ellipse is 5x² + y² = 5.

Given that we need to find the equation of the ellipse whose length of major axis is 26 and foci (±5,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that length of the major axis is 2a

⇒ 2a = 26

⇒ a = 13

⇒ a² = 169

We know that foci = (±ae,0)

⇒ ae = 5

⇒ 13e = 5

⇒

We know that eccentricity

⇒

⇒

⇒

⇒ b² = 144

The equation of the ellipse is

⇒

∴ The equation of the ellipse is .

Given that we need to find the equation of the ellipse whose length of the minor axis is 16 and foci (0,±6).

Let us assume the equation of the ellipse as (b²>a²).

We know that the length of the minor axis of the ellipse is 2a,

⇒ 2a = 16

⇒ a = 8

⇒ a² = 64

We know that foci = (0,±be)

⇒ be = 6

We know that eccentricity

⇒

⇒ 36 = b² - 64

⇒ b² = 100

The equation of the ellipse is

⇒

∴ The equation of the ellipse is .

Given that we need to find the equation of the ellipse whose foci are (±3,0) and a = 4.

Let us assume the equation of the ellipse as (a²>b²).

⇒ a = 4

⇒ a² = 16

We know that foci = (±ae,0)

⇒ ae = 3

⇒ 4e = 3

⇒

We know that eccentricity

⇒

⇒

⇒

⇒ b² = 7

The equation of the ellipse is

⇒

⇒

⇒ 7x² + 16y² = 112

∴ The equation of the ellipse is 7x² + 16y² = 112.

Given that we need to find the equation of the ellipse whose eccentricity is and foci (±4,0).

Let us assume the equation of the ellipse as (a²>b²).

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

We know that foci = (±ae,0)

⇒ ae = 4

⇒

⇒ a = 12

⇒ a² = 144

⇒

⇒ b² = 128

The equation of the ellipse is

⇒

∴ The equation of the ellipse is .

Given that we need to find the equation of the ellipse whose minor axis is equal to the distance between foci and length of latus rectum is 10.

Let us assume the equation of the ellipse as (a²>b²).

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

We know that eccentricity

⇒ 2b = 2ae

⇒ b = ae

⇒

⇒ b² = a² - b²

⇒ a² = 2b² ..... - - - - (1)

We know that the length of the latus rectum is .

⇒

From (1)

⇒

⇒ a = 10

⇒ a² = 100

⇒

⇒ b² = 50

The equation of the ellipse is

⇒

⇒

⇒ x² + 2y² = 100

∴ The equation of the ellipse is x² + 2y² = 100.

Given that we need to find the equation of the ellipse whose centre is (- 2,3) and whose semi - axis are 3 and 2.

(i) If major axis is parallel to the x - axis.

We know that the equation of the ellipse with centre (p,q) is given by .

Since major axis is parallel to x - axis a²>b².

So, a = 3 and b = 2.

⇒ a² = 9

⇒ b² = 4

The equation of the ellipse is

⇒

⇒

⇒ 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) If major axis is parallel to the y - axis.

We know that the equation of the ellipse with centre (p,q) is given by .

Since major axis is parallel to y - axis b²>a².

So, a = 2 and b = 3.

⇒ a² = 4

⇒ b² = 9

The equation of the ellipse is

⇒

⇒

⇒ 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.

Given that 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 .

⇒

⇒ a = 2b .... (1)

We know that eccentricity of an ellipse is

From (1)

⇒

⇒

⇒

⇒

⇒ .

(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 .

⇒

⇒ a² = 2b² .... (1)

We know that eccentricity of an ellipse is

From (1)

⇒

⇒

⇒

⇒

⇒ .

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse x² + 2y² - 2x + 12y + 10 = 0.

⇒ x² + 2y² - 2x + 12y + 10 = 0

⇒ (x² - 2x + 1) + 2(y² + 6y + 9) - 9 = 0

⇒ (x - 1)² + 2(y + 3)² = 9

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (1, - 3)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

Length of the major axis 2a = 2(3) = 6

Length of the minor axis 2b = = 3

⇒ Foci = (p±ae,q)

⇒ Foci =

⇒ Foci =

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse x² + 4y² - 4x + 24y + 31 = 0.

⇒ x² + 4y² - 4x + 24y + 31 = 0

⇒ (x² - 4x + 4) + 4(y² + 6y + 9) - 9 = 0

⇒ (x - 2)² + 4(y + 3)² = 9

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (2, - 3)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

Length of the major axis 2a = 2(3) = 6

Length of the minor axis 2b = = 3

⇒ Foci = (p±ae,q)

⇒ Foci =

⇒ Foci =

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse 4x² + y² - 8x + 2y + 1 = 0.

⇒ 4x² + y² - 8x + 2y + 1 = 0

⇒ 4(x² - 2x + 1) + (y² + 2y + 1) - 4 = 0

⇒ 4(x - 1)² + (y + 1)² = 4

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (1, - 1)

Here b²>a²

⇒ eccentricity(e) =

⇒

⇒

⇒

Length of the major axis 2b = 2(2) = 4

Length of the minor axis 2a = 2(1) = 2

⇒ Foci = (p,q±be)

⇒ Foci =

⇒ Foci =

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse 3x² + 4y² - 12x - 8y + 4 = 0.

⇒ 3x² + 4y² - 12x - 8y + 4 = 0

⇒ 3(x² - 4x + 4) + 4(y² - 2y + 1) - 12 = 0

⇒ 3(x - 2)² + 4(y - 1)² = 12

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (2,1)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

Length of the major axis 2a = 2(2) = 4

Length of the minor axis 2b = = 2

⇒ Foci = (p±ae,q)

⇒ Foci =

⇒ Foci =

⇒ Foci = (3,1) and (1,1)

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse 4x² + 16y² - 24x - 32y - 120 = 0.

⇒ 4x² + 16y² - 24x - 32y - 120 = 0

⇒ 4(x² - 6x + 9) + 16(y² - 2y + 1) - 172 = 0

⇒ 4(x - 3)² + 16(y - 1)² = 172

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (3,1)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

Length of the major axis 2a = 2() =

Length of the minor axis 2b =

⇒ Foci = (p±ae,q)

⇒ Foci =

⇒ Foci =

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse x² + 4y² - 2x = 0.

⇒ x² + 4y² - 2x = 0

⇒ (x² - 2x + 1) + 4(y²) - 1 = 0

⇒ (x - 1)² + 4(y - 0)² = 1

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (1,0)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

Length of the major axis 2a = 2(1) = 2

Length of the minor axis 2b = = 1

⇒ Foci = (p±ae,q)

⇒ Foci =

⇒ Foci =

Given that we need to find the equation of the ellipse whose foci are at (±4,0) and passes through (4,1).

Let us assume the equation of the ellipse is - - - - (1) (a²>b²).

We know that foci are (±ae,0) and eccentricity of the ellipse is

⇒ ae = 3

⇒

⇒ a² - b² = 9 ..... - - - (2)

Substituting the point (4,1) in (1) we get,

⇒

⇒

⇒ 16b² + a² = a²b²

From (2),

⇒ 16(a² - 9) + a² = a²(a² - 9)

⇒ 16a² - 144 + a² = a⁴ - 9a²

⇒ a⁴ - 26a² + 144 = 0

⇒ a⁴ - 18a² - 8a² + 144 = 0

⇒ a²(a² - 18) - 8(a² - 18) = 0

⇒ (a² - 8)(a² - 18) = 0

⇒ a² - 8 = 0 (or) a² - 18 = 0

⇒ a² = 8 (or) a² = 18

⇒ b² = 18 - 9(since b²>0)

⇒ b² = 9.

The equation of the ellipse is

⇒

⇒

⇒ x² + 2y² = 18

∴ The equation of the ellipse is x² + 2y² = 18.

Given that we need to find the equation of the ellipse whose eccentricity is , latus - rectum is 5 and centre is at origin.

Let us assume the equation of the ellipse is - - - - (1) (a²>b²) since centre is at origin.

We know that eccentricity of the ellipse is

⇒

⇒

⇒ 9(a² - b²) = 4a²

⇒ 5a² = 9b²

⇒ ..... - - - (2)

We know that length of the latus - rectum is

⇒

⇒

⇒

⇒

⇒

From (2),

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒

⇒ 20x² + 36y² = 405

∴ The equation of the ellipse is 20x² + 36y² = 405.

Given that we need to find the equation of the ellipse whose eccentricity is , centre at the origin and passes through (6,4).

Let us assume the equation of the ellipse is - - - - (1) (a²<b²), since centre is at origin and foci on y - axis.

We know that eccentricity of the ellipse is

⇒

⇒

⇒ 16b² - 16a² = 9b²

⇒ 7b² = 16a²

⇒ ..... - - - (2)

Substituting the point (6,4) in (1) we get,

⇒

⇒

⇒

⇒

⇒ 7b² = 688

⇒

From (2),

⇒

⇒

⇒ a² = 43

The equation of the ellipse is

⇒

⇒

⇒

⇒ 16x² + 7y² = 688

∴ The equation of the ellipse is 16x² + 7y² = 688.

Given that we need to find the equation of the ellipse passing through the points (4,3) and (- 1,4).

Let us assume the equation of the ellipse as (a²>b²). - - - - (1)

Substituting the point (4,3) in (1) we get

⇒

⇒

⇒

⇒ 16b² + 9a² = a² b² ..... - - (2)

Substituting the point (- 1,4) in (1) we get

⇒

⇒

⇒

⇒ b² + 16a² = a²b² ..... - - (3)

(3)×16 - (2)

⇒ (16b² + 256a²) - (9a² + 16b²) = (16a²b² - a²b²)

⇒ 247a² = 15a²b²

⇒ 15b² = 247

⇒

From (3)

⇒

⇒

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒ 7x² + 15y² = 247

∴ The equation of the ellipse is 7x² + 15y² = 247.

Given that we need to find the equation of the ellipse whose eccentricity is and passes through (- 3,1).

Let us assume the equation of the ellipse is - - - - (1) (a²>b²).

We know that eccentricity of the ellipse is

⇒

⇒

⇒ 5a² - 5b² = 2a²

⇒ 5b² = 3a²

⇒ ..... - - - (2)

Substituting the point (- 3,1) in (1) we get,

⇒

⇒

⇒

⇒

⇒ 5b² = 32

⇒

From (2),

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒

⇒ 3x² + 5y² = 32

∴ The equation of the ellipse is 3x² + 5y² = 32.

Given that we need to find the equation of the ellipse whose distance between the foci is 8 units and distance between the directrices is 18 units.

We know that the distance between the foci is 2ae.

⇒ 2ae = 8

⇒ ae = 4 ..... - (1)

We know that the distance between the directrices is .

⇒

⇒ ..... - (2)

(1)×(2)

⇒

⇒ a² = 36 .... (3)

⇒ a = 6

We know that eccentricity of an ellipse is

⇒

⇒ 36 - b² = 16

⇒ b² = 20 ..... (4)

The equation of the ellipse is,

⇒

⇒

⇒ 5x² + 9y² = 180

∴ The equation of the ellipse is 5x² + 9y² = 180.

Given that we need to find the equation of the ellipse whose vertices are (0,±10) and eccentricity .

Let us assume the equation of the ellipse as - - - - (1) (a²>b²).

We know that vertices of the ellipse are (0,±b)

⇒ b = 10

⇒ b² = 100

We know that eccentricity

⇒

⇒

⇒

⇒

The equation of the ellipse is

⇒

⇒

⇒

⇒ 9x² + 25y² = 2500

∴ The equation of the ellipse is 9x² + 25y² = 2500.

Given that we need to find the locus of the point on the rod whose ends always touching the coordinate axes.

We need to the equation of locus of point P on the rod, which is 3 cm from the end in contact with x - axis.

Let us assume AB be the rod of length 12 cm and P(x,y) be the required point.

From the figure using similar triangles DAP and CBP we get,

⇒

⇒

⇒ q = 3y ..... (1)

⇒

⇒

⇒ ..... - (2)

Now OB = OC + CB

⇒

⇒ .... (3)

⇒ OA = OD + DA

⇒ OA = y + 3y

⇒ OA = 4y .... (4)

Since OAB is a right angled triangle,

⇒ OA² + OB² = AB²

⇒

⇒

⇒

⇒

⇒ x² + 9y² = 81

∴ The equation of the ellipse is x² + 9y² = 81.

Given that we need to find the equation of set of all points whose distances from S(0,4) are of their distances from the line(M) y = 9.

Let P(x,y) be any point from the set of all points.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒

⇒

⇒

⇒

⇒

⇒ 9x² + 9y² - 72y + 144 = 4y² - 72y + 324

⇒ 9x² + 5y² = 180

∴ The equation of the ellipse is 9x² + 5y² = 180.

Given that we need to find the equation of the ellipse whose semi - major and semi - minor axes are 2 and and their corresponding equations of equations of axes are y - 5 = 0 and x + 3 = 0.

We know that the centre is the point of intersection of both axes. So, on solving these axes we get the centre to be (- 3,5).

⇒ Semi - major axis(a) = 2

⇒ a² = 4

⇒ Semi - minor axis(b) =

⇒ b² = 3

We know that the equation of the ellipse whose centre is (p,q) and the length of semi - major axis is a and semi - minor axis is b is .

∴ The equation of the ellipse is .

Given the equation of the ellipse is 9x² + 5y² - 18x - 2y - 16 = 0.

We need to find the eccentricity.

Given equation can be rewritten as

⇒

⇒

⇒

⇒ .

We know for the ellipse (b²>a²)

⇒

Here and , b²>a²

⇒

⇒

⇒

⇒

∴ The eccentricity is .

Given that we need to find the centre and eccentricity of the ellipse 3x² + 4y² - 6x + 8y - 5 = 0.

⇒ 3x² + 4y² - 6x + 8y - 5 = 0

⇒ 3(x² - 2x + 1) + 4(y² + 2y + 1) = 12

⇒ 3(x - 1)² + 4(y + 1)² = 12

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (1, - 1)

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

Given that PSQ is the focal chord of the ellipse 4x² + 9y² = 36.

It is also given that SP = 4. We need to find the value of S’Q, where S and S’ are foci.

Given ellipse is rewritten as .

We got a² = 9, a = 3 and b² = 4.

We know that semi latus rectum is the harmonic mean of any two segments of focal chord.

⇒

⇒

⇒

⇒

We know that SQ + S’Q = 2a

⇒

⇒

⇒

Given that we need to find the eccentricity of the ellipse whose latus - rectum is one half of the minor axis.

We know that length of latus rectum is and length of minor axis is 2b.

⇒

⇒ a = 2b.

We know that eccentricity of the ellipse is

⇒

⇒

⇒ .

Given that we need to find the eccentricity of the ellipse whose latus - rectum is equal to the distance between the foci.

We know that length of latus rectum is and distance between foci is 2ae.

⇒

⇒ b² = a²e.

We know that eccentricity of the ellipse is b² = a²(1 - e²)

⇒ a²(1 - e²) = a²e

⇒ e² + e - 1 = 0

⇒

⇒ (since e>0)

Given that S and S’ are the foci of the ellipse .

It is told that ΔBSS’ is equilateral, where B is the end of the minor axis. We need to find the eccentricity of the ellipse.

Let us assume that B = (0,b)

We know that foci of the ellipse are (±ae,0).

We know that the distance between the foci is 2ae.

Let us find the distance SB

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

⇒

⇒ .

We know that sides of an equilateral triangle are equal.

⇒ SB = 2ae

⇒ SB² = 4a²e²

⇒ a²e² + b² = 4a²e²

We know that b² = a²(1 - e²),

⇒ a²e² + a² - a²e² = 4a²e²

⇒ a² = 4a²e²

⇒

⇒

⇒ .

Given that the minor axis of an ellipse subtends an equilateral triangle with vertex at one end of major axis. We need to find the eccentricity of the ellipse.

Let us assume that ends of minor axis be B = (0,b) and C(0, - b) and end of major axis be A(a,0)

We know that the distance between the ends of minor axis is 2b.

Let us find the distance AB

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

⇒

⇒ .

We know that sides of an equilateral triangle are equal.

⇒ AB = 2b

⇒ AB² = 4b²

⇒ a² + b² = 4b²

We know that b² = a²(1 - e²),

⇒ a² + a² - a²e² = 4a² - 4a²e²

⇒ 2a² = 3a²e²

⇒

⇒ .

Given that the latus rectum of an ellipse subtends an right angle with centre of the ellipse. We need to find the eccentricity of the ellipse.

Let us assume that the equation of the ellipse be (a²>b²) such that the centre is O(0,0).

We know that the ends A and B of the latus rectum are .

Let us find the slope(m₁) of the OA.

We know that the slope of the line joining points (x₁,y₁) and (x₂,y₂) is

⇒

⇒

Let us find the slope(m₂) of the OB.

We know that the slope of the line joining points (x₁,y₁) and (x₂,y₂) is

⇒

⇒

We know that the product of the slopes of the perpendicular is - 1.

⇒ m₁.m₂ = - 1

⇒ = - 1

⇒ b⁴ = a⁴e²

⇒

⇒

Given that we need to find the centre, lengths of axes, eccentricity and foci of the ellipse 12x² + 4y² + 24x - 16y + 25 = 0.

⇒ 12x² + 4y² + 24x - 16y + 25 = 0

⇒ 12(x² + 2x + 1) + 4(y² - 4y + 4) - 3 = 0

⇒ 12(x + 1)² + 4(y - 2)² = 3

⇒

⇒

Comparing with the standard form

⇒ Centre = (p,q) = (1, - 2)

Here b²>a²

⇒ eccentricity(e) =

⇒

⇒

⇒

Length of the major axis 2b =

Length of the minor axis 2a = = 1

∴ The correct option is D

Given that we need to find the equation of the ellipse whose focus is S(- 1,1) and directrix(M) is x - y + 3 = 0 and eccentricity(e) is equal to .

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

We know that the distance between the focus and any point on ellipse is equal to the eccentricity times the perpendicular distance from that point to the directrix.

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

We know that the perpendicular distance from the point (x₁,y₁) to the line ax + by + c = 0 is .

⇒ SP = ePM

⇒ SP² = e²PM²

⇒

⇒

⇒

⇒ 8x² + 8y² + 16x - 16y + 16 = x² + y² - 2xy - 6y + 6x + 9

⇒ 7x² + 7y² + 2xy + 10x - 10y + 7 = 0

∴ The correct option is B

Given that we need to find the equation of the circle whose end points of diameter are the foci of ellipse .

We know that foci of ellipse are (±ae,0).

We know that (0,0) is the centre of the ellipse and midpoint of the foci.

We know that distance between centre and any focus is ae.

So, we have with centre at (0,0) and radius ae.

We know that eccentricity of the ellipse

We know that the equation of the circle whose centre is (p,q) and radius ‘r’ is (x - p)² + (y - q)² = r²

⇒ (x - 0)² + (y - 0)² = (ae)²

⇒ x² + y² = a²e²

⇒

⇒ x² + y² = a² - b²

∴ The correct option is D

Given that we need to find the eccentricity of the ellipse whose latus - rectum is one half of the minor axis.

We know that length of latus rectum is and length of minor axis is 2b.

⇒

⇒ a = 2b.

We know that eccentricity of the ellipse is

⇒

⇒

⇒

∴ The correct option is A

Given that we need to find the eccentricity of the ellipse whose latus - rectum is equal to the distance between the foci.

We know that length of latus rectum is and distance between foci is 2ae.

⇒

⇒ b² = a²e.

We know that eccentricity of the ellipse is b² = a²(1 - e²)

⇒ a²(1 - e²) = a²e

⇒ e² + e - 1 = 0

⇒

⇒ (since e>0)

∴ The correct option is A

Given that we need to find the eccentricity of the ellipse whose minor axis is equal to the distance between the foci.

We know that distance between foci is 2ae and length of minor axis is 2b.

⇒ 2ae = 2b

⇒ b = ae

⇒ b² = a²e²

We know that b² = a²(1 - e²)

⇒ a²(1 - e²) = a²e²

⇒ a² = 2a²e²

⇒

⇒

∴ The correct option is C

Given that we need to find the difference between the lengths of major axis and length of latus rectum of an ellipse.

We know that the length of major axis is 2a and latus rectum is for the ellipse .

Let d be the difference.

⇒

⇒

We know that b² = a²(1 - e²)

⇒

⇒

⇒ d = 2ae²

∴ The correct option is D

Given that we need to find the eccentricity of the conic 9x² + 25y² = 225.

It is rewritten as

We know that the eccentricity of ellipse is (a²>b²).

⇒

⇒

⇒

∴ The correct option is B

Given conic is 3x² + 4y² - 6x + 8y - 5 = 0. It is re written as

⇒ 3x² + 4y² - 6x + 8y - 5 = 0

⇒ 3(x² - 2x + 1) + 4(y² + 2y + 1) = 12

⇒ 3(x - 1)² + 4(y + 1)² = 12

⇒

⇒

Comparing with (a²>b²), The length of the latus - rectum is .

⇒

⇒

∴ The correct option is A

Given that we need to find the equation of the tangents to the ellipse 9x² + 16y² = 144 from the point (2,3).

We know that tangent at any point (x₁,y₁) on the ellipse is S₁ = 0.

⇒ S₁ = 0

⇒ 9(xx₁) + 16(yy₁) = 144 .... (1)

This passes through the point (2,3)

⇒ 9(2x₁) + 16(3y₁) = 144

⇒ 18x₁ + 48y₁ = 144

⇒ 3x₁ + 8y₁ = 24

⇒ 8y₁ = 24 - 3x₁

⇒ .... - - (2)

Substituting this in the equation of the ellipse we get,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

From (2)

⇒

⇒ y₁ = 3

⇒

⇒

⇒

Substituting x₁ = 0 and y₁ = 3 in (1), we get

⇒ 9(x(0)) + 16(y(3)) = 144

⇒ 48y = 144

⇒ y = 3.

Substituting and in (1), we get

⇒

⇒

⇒ x + y = 5

∴ The correct option is D

Given that we need to find the eccentricity of the ellipse 4x² + 9y² + 8x + 36y + 4 = 0.

⇒ 4x² + 9y² + 8x + 36y + 4 = 0

⇒ 4(x² + 2x + 1) + 9(y² + 4y + 4) - 36 = 0

⇒ 4(x + 1)² + 9(y + 2)² = 36

⇒

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

∴ The correct option is D

Given that we need to find the eccentricity of the ellipse 4x² + 9y² = 36.

⇒ 4x² + 9y² = 36

⇒

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

∴ The correct option is C

Given that we need to find the eccentricity of the ellipse 5x² + 9y² = 1.

⇒ 5x² + 9y² = 1

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

∴ The correct option is A

Given ellipse is x² + 4y² = 9.

⇒ x² + 4y² = 9

⇒

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

Length of the latus rectum is

Focus (±ae,0) =

Directrices are

Directrices are

∴ The correct options are B and D

Given that we need to find the eccentricity of the ellipse whose latus - rectum is one half of the minor axis.

We know that length of latus rectum is and length of minor axis is 2b.

⇒

⇒ a = 2b.

We know that eccentricity of the ellipse is

⇒

⇒

⇒ .

∴ The correct option is C

Given that we need to find the equation of the ellipse whose centre is (1, - 1) and semi - major axis 8 and passes through (1,3).

From the standard equation .

Centre = (p,q) = (1, - 1)

a = 8

a² = 64

The equation is . This passes through (1,3).

Substituting in the curve, we get,

⇒

⇒

⇒ b² = 16

The equation of the ellipse is .

∴ The correct option is B

Given ellipse is 9x² + 16y² = 144. It is rewritten as

⇒

⇒

Comparing with standard equation we get

⇒ a² = 16

⇒ a = 4

We know that the sum of the focal distances of any point on the ellipse is 2a.

⇒ 2(4) = 8

∴ The correct option is D

Given (2,4) and (10,10) are the ends of the latus - rectum and eccentricity is .

We know that length of the latus rectum is .

We know that the distance between the two points (x₁,y₁) and (x₂,y₂) is .

⇒

⇒

⇒

⇒

⇒ 2b² = 10a

⇒ b² = 5a

We know that b² = a²(1 - e²)

⇒ a²(1 - e²) = 5a

⇒

⇒

⇒

⇒

∴ The correct option is A

Given equation is

⇒

⇒

Comparing with standard equation we get

⇒ λ - 2>0

⇒ λ>2 .... - (1)

⇒ 5 - λ>0

⇒ λ<5 .... - (2)

From (1) and (2),

⇒ 2<λ<5

∴ The correct option is C

Given that we need to find the eccentricity of the ellipse 9x² + 25y² - 18x - 100y - 116 = 0.

⇒ 9x² + 25y² - 18x - 100y - 116 = 0

⇒ 9(x² - 2x + 1) + 25(y² - 4y + 4) - 225 = 0

⇒ 9(x - 1)² + 25(y - 2)² = 225

⇒

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

∴ The correct option is B

Given that the major axis of an ellipse is three times the minor axis.

⇒ a = 3b

We know that eccentricity(e) =

⇒

⇒

⇒

⇒

∴ The correct option is D

Given that we need to find the eccentricity of the ellipse 25x² + 16y² = 400.

⇒ 25x² + 16y² = 400

⇒

⇒

Comparing with the standard form

Here b²>a²

⇒ eccentricity(e) =

⇒

⇒

⇒

∴ The correct option is A

Given that we need to find the eccentricity of the ellipse 5x² + 9y² = 1.

⇒ 5x² + 9y² = 1

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

⇒

∴ The correct option is A

Given that we need to find the eccentricity of the ellipse 4x² + 9y² = 36.

⇒ 4x² + 9y² = 36

⇒

⇒

Comparing with the standard form

Here a²>b²

⇒ eccentricity(e) =

⇒

⇒

⇒

∴ The correct option is C