Ellipse

Given:

…(i)

Since, 25 > 9

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 25 and b² = 9

⇒ a = √25 and b = √9

⇒ a = 5 and b = 3

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 5

= 10 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (5, 0) and (-5, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 25 – 9

c² = 16

c = √16

c = 4 …(I)

∴ Coordinates of foci = (±4, 0)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

…(i)

Since, 49 > 36

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 49 and b² = 36

⇒ a = √49 and b = √36

⇒ a = 7 and b = 6

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 7

= 14 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (7, 0) and (-7, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 49 – 36

c² = 13

c = √13 …(I)

∴ Coordinates of foci = (±√13, 0)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

16x² + 25y² = 400

Divide by 400 to both the sides, we get

…(i)

Since, 25 > 4

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 25 and b² = 4

⇒ a = √25 and b = √4

⇒ a = 5 and b = 2

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 5

= 10 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (5, 0) and (-5, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 25 – 4

c² = 21

c = √21 …(I)

∴ Coordinates of foci = (±√21, 0)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

x² + 4y² = 100

Divide by 100 to both the sides, we get

…(i)

Since, 100 > 25

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 100 and b² = 25

⇒ a = √100 and b = √25

⇒ a = 10 and b = 5

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 10

= 20 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (10, 0) and (-10, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 100 – 25

c² = 75

c = √75

c = 5√3 …(I)

∴ Coordinates of foci = (±5√3, 0)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

9x² + 16y² = 144

Divide by 144 to both the sides, we get

…(i)

Since, 16 > 9

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 16 and b² = 9

⇒ a = √16 and b = √9

⇒ a = 4 and b = 3

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 2 × 4

= 8 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

= (4, 0) and (-4, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 16 – 9

c² = 7

c = √7 …(I)

∴ Coordinates of foci = (±√7, 0)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

4x² + 9y² = 1

…(i)

Since,

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

(i) To find: Length of major axes

Clearly, a > b, therefore the major axes of the ellipse is along x axes.

∴Length of major axes = 2a

= 1 unit

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (a, 0) and (-a, 0)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (±c, 0) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

…(I)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

…(i)

Since, 4 < 25

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 25 and b² = 4

⇒ a = √25 and b = √4

⇒ a = 5 and b = 2

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 5

= 10 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 5) and (0, -5)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 25 – 4

c² = 21

c = √21 …(I)

∴ Coordinates of foci = (0, ±√21)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

…(i)

Since, 9 < 16

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 16 and b² = 9

⇒ a = √16 and b = √9

⇒ a = 4 and b = 3

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 4

= 8 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 4) and (0, -4)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 16 – 9

c² = 7

c = √7 …(I)

∴ Coordinates of foci = (0, ±√7)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

3x² + 2y² = 18

Divide by 18 to both the sides, we get

…(i)

Since, 6 < 9

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 9 and b² = 6

⇒ a = √9 and b = √6

⇒ a = 3 and b = √6

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 3

= 6 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 6) and (0, -6)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 9 – 6

c² = 3

c = √3 …(I)

∴ Coordinates of foci = (0, ±√3)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

= 4

Given:

9x² + y² = 36

Divide by 36 to both the sides, we get

…(i)

Since, 4 < 36

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 36 and b² = 4

⇒ a = √36 and b = √4

⇒ a = 6 and b = 2

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 6

= 12 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 6) and (0, -6)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 36 – 4

c² = 32

c = √32 …(I)

∴ Coordinates of foci = (0, ±√32)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

16x² + y² = 16

Divide by 16 to both the sides, we get

…(i)

Since, 1 < 16

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 16 and b² = 1

⇒ a = √16 and b = √1

⇒ a = 4 and b = 1

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 4

= 8 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 4) and (0, -4)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 16 – 1

c² = 15

c = √15 …(I)

∴ Coordinates of foci = (0, ±√15)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given:

25x² + 4y² = 100

Divide by 100 to both the sides, we get

…(i)

Since, 4 < 25

So, above equation is of the form,

…(ii)

Comparing eq. (i) and (ii), we get

a² = 25 and b² = 4

⇒ a = √25 and b = √4

⇒ a = 5 and b = 2

(i) To find: Length of major axes

Clearly, a < b, therefore the major axes of the ellipse is along y axes.

∴Length of major axes = 2a

= 2 × 5

= 10 units

(ii) To find: Coordinates of the Vertices

Clearly, a > b

∴ Coordinate of vertices = (0, a) and (0, -a)

= (0, 5) and (0, -5)

(iii) To find: Coordinates of the foci

We know that,

Coordinates of foci = (0, ±c) where c² = a² – b²

So, firstly we find the value of c

c² = a² – b²

= 25 – 4

c² = 21

c = √21 …(I)

∴ Coordinates of foci = (0, ±√21)

(iv) To find: Eccentricity

We know that,

[from (I)]

(v) To find: Length of the Latus Rectum

We know that,

Given: Vertices = (±6, 0) …(i)

The vertices are of the form = (±a, 0) …(ii)

Hence, the major axis is along x – axis

∴ From eq. (i) and (ii), we get

a = 6

⇒ a² = 36

and We know that, if the major axis is along x – axis then the equation of Ellipse is of the form of

Also, given coordinate of foci = (±4, 0) …(iii)

We know that,

Coordinates of foci = (±c, 0) …(iv)

∴ From eq. (iii) and (iv), we get

c = 4

We know that,

c² = a² – b²

⇒ (4)² = (6)² – b²

⇒ 16 = 36 – b²

⇒ b² = 36 – 16

⇒ b² = 20

Substituting the value of a² and b² in the equation of an ellipse, we get

Given: Vertices = (0, ±4) …(i)

The vertices are of the form = (0, ±a) …(ii)

Hence, the major axis is along y – axis

∴ From eq. (i) and (ii), we get

a = 4

⇒ a² = 16

and We know that, if the major axis is along y – axis then the equation of Ellipse is of the form of

Also, given coordinate of foci = (0, ±√7) …(iii)

We know that,

Coordinates of foci = (0, ±c) …(iv)

∴ From eq. (iii) and (iv), we get

c = √7

We know that,

c² = a² – b²

⇒ (√7)² = (4)² – b²

⇒ 7 = 16 – b²

⇒ b² = 16 – 7

⇒ b² = 9

Substituting the value of a² and b² in the equation of an ellipse, we get

Given:

Ends of Major Axis = (±4, 0)

and Ends of Minor Axis = (0, ±3)

Here, we can see that the major axis is along the x – axis.

∴ The Equation of Ellipse is of the form,

…(i)

where, a is the semi – major axis and b is the semi – minor axis.

Accordingly, a = 4 and b = 3

Substituting the value of a and b in eq. (i), we get

Let the equation of the required ellipse be

Given: Length of Major Axis = 20units …(i)

We know that,

Length of Major Axis = 2a …(ii)

∴ From eq. (i) and (ii), we get

2a = 20

⇒ a = 10

It is also given that,

Coordinates of foci = (±5√3, 0) …(iii)

We know that,

Coordinates of foci = (±c, 0) …(iv)

∴ From eq. (iii) and (iv), we get

c = 5√3

We know that,

c² = a² – b²

⇒ (5√3)² = (10)² – b²

⇒ 75 = 100 – b²

⇒ b² = 100 – 75

⇒ b² = 25

Substituting the value of a² and b² in the equation of an ellipse, we get

Let the equation of the required ellipse be

Given:

Coordinates of foci = (±2, 0) …(iii)

We know that,

Coordinates of foci = (±c, 0) …(iv)

∴ From eq. (iii) and (iv), we get

c = 2

It is also given that

we know that,

[∵ c = 2]

⇒ a = 4

Now, we know that,

c² = a² – b²

⇒ (2)² = (4)² – b²

⇒ 4 = 16 – b²

⇒ b² = 16 – 4

⇒ b² = 12

Substituting the value of a² and b² in the equation of an ellipse, we get

Let the equation of the required ellipse be

Given:

Coordinates of foci = (±1, 0) …(i)

We know that,

Coordinates of foci = (±c, 0) …(ii)

∴ From eq. (i) and (ii), we get

c = 1

It is also given that

we know that,

[∵ c = 1]

⇒ a = 2

Now, we know that,

c² = a² – b²

⇒ (1)² = (2)² – b²

⇒ 1 = 4 – b²

⇒ b² = 4 – 1

⇒ b² = 3

Substituting the value of a² and b² in the equation of an ellipse, we get

Given:

Coordinates of foci = (0, ±4) …(i)

We know that,

Coordinates of foci = (0, ±c) …(ii)

The coordinates of the foci are (0, ±4). This means that the major and minor axes are along y and x axes respectively.

∴ From eq. (i) and (ii), we get

c = 4

It is also given that

we know that,

[∵ c = 4]

⇒ a = 5

Now, we know that,

c² = a² – b²

⇒ (4)² = (5)² – b²

⇒ 16 = 25 – b²

⇒ b² = 25 – 16

⇒ b² = 9

Since, the foci of the ellipse are on y – axis. So, the Equation of Ellipse is

Substituting the value of a² and b² , we get

Given: Center is at the origin

and Major axis is along x – axis

So, Equation of ellipse is of the form

…(i)

Given that ellipse passing through the points (4, 3) and (-1, 4)

So, point (4, 3) and (-1, 4) will satisfy the eq. (i)

Taking point (4, 3) where x = 4 and y = 3

Putting the values in eq. (i), we get

…(ii)

Taking point (-1, 4) where x = -1 and y = 4

Putting the values in eq. (i), we get

…(iii)

Now, we have to solve the above two equations to find the value of a and b

Multiply the eq. (iii) by 16, we get

…(iv)

Subtracting eq. (iv) from (ii), we get

Substituting the value of b² in eq. (iii), we get

Thus,

Substituting the value of a² and b² in eq. (i), we get

⇒ 7x² + 15y² = 247

Given that

we know that,

We know that,

c² = a² – b²

…(i)

It is also given that Coordinates of foci is on the y – axis

So, Equation of ellipse is of the form

Substituting the value of b² in above eq., we get

…(ii)

Given that ellipse passing through the points (6, 4)

So, point (6, 4) will satisfy the eq. (ii)

Taking point (6, 4) where x = 6 and y = 4

Putting the values in eq. (ii), we get

Substituting the value of a² in eq. (i), we get

Substituting the value of a² and b² in the equation of an ellipse, we get

or 16x² + 7y² = 688

Let the equation of the required ellipse be

…(i)

Given:

Coordinates of foci = (±3, 0) …(ii)

We know that,

Coordinates of foci = (±c, 0) …(iii)

∴ From eq. (ii) and (iii), we get

c = 3

We know that,

c² = a² – b²

⇒ (3)² = a² – b²

⇒ 9 = a² – b²

⇒ b² = a² – 9 …(iv)

Given that ellipse passing through the points (4, 1)

So, point (4, 1) will satisfy the eq. (i)

Taking point (4, 1) where x = 4 and y = 1

Putting the values in eq. (i), we get

[from (iv)]

⇒ 16a² – 144 + a² = a²(a² – 9)

⇒ 17a² – 144 = a⁴ – 9a²

⇒ a⁴ – 9a² – 17a² + 144 = 0

⇒ a⁴ – 26a² + 144 = 0

⇒ a⁴ – 8a² – 18a² + 144 = 0

⇒ a²(a² – 8) – 18(a² – 8) = 0

⇒ (a² – 8)(a² – 18) = 0

⇒ a² – 8 = 0 or a² – 18 = 0

⇒ a² = 8 or a² = 18

If a² = 8 then

b² = 8– 9

= - 1

Since the square of a real number cannot be negative. So, this is not possible

If a² = 18 then

b² = 18 – 9

= 9

So, equation of ellipse if a² = 18 and b² = 9

Let the equation of required ellipse is

…(A)

Given:

Length of Major Axis = 10units …(i)

We know that,

Length of major axis = 2a …(ii)

∴From eq. (i) and (ii), we get

2a = 10

⇒ a = 5

It is also given that

Length of Minor Axis = 8 units …(iii)

We know that,

Length of minor axis = 2b …(iv)

∴From eq. (iii) and (iv), we get

2b = 8

⇒ b = 4

Substituting the value of a and b in eq. (A), we get

Let the equation of the required ellipse is

…(i)

Given that

we know that,

We know that,

c² = a² – b²

…(ii)

It is also given that, Latus Rectum = 5 …(iii)

We know that,

Substituting the value of a in eq. (ii), we get

Substituting the value of a² and b² in eq. (i), we get

Let the equation of the required ellipse is

…(i)

It is given that,

We know that,

and Length of Minor Axis = 2b

So, according to the given condition,

⇒ 2b = a …(ii)

Now, we have to find the eccentricity

We know that,

…(iii)

where, c² = a² – b²

So, c² = (2b)² – b² [from (ii)]

⇒ c² = 4b² – b²

⇒ c² = 3b²

⇒ c = √3b²

⇒ c = b√3

Substituting the value of c and a in eq. (iii), we get

Let the equation of the required ellipse is

…(i)

It is given that,

We know that,

and Length of Minor Axis = 2a

So, according to the given condition,

⇒ 2b² = a² …(ii)

⇒ a = √2b²

⇒a = b√2

Now, we have to find the eccentricity

We know that,

…(iii)

where, c² = a² – b²

So, c² = 2b² – b² [from (ii)]

⇒ c² = b²

⇒ c = √b²

⇒ c = b

Substituting the value of c and a in eq. (iii), we get