Question - 1 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/36 + y²/16= 1

Answer - 1 : -

Given:

The equation is x²/36+ y²/16 = 1

Here, the denominator of x²/36 isgreater than the denominator of y²/16.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x²/a² +y²/b² =1, we get

a = 6 and b = 4.

c = √(a² – b²)

= √(36-16)

= √20

= 2√5

Then,

The coordinates of the foci are (2√5, 0) and (-2√5, 0).

The coordinates of the vertices are (6, 0) and (-6, 0)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, e = c/a =2√5/6 = √5/3

Length of latus rectum = 2b²/a= (2×16)/6 = 16/3

Question - 2 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/4 + y²/25= 1

Answer - 2 : -

Given:

The equation is x²/4 +y²/25 = 1

Here, the denominator of y²/25 isgreater than the denominator of x²/4.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x²/a² + y²/b² = 1, we get

a = 5 and b = 2.

c = √(a² – b²)

= √(25-4)

= √21

Then,

The coordinates of the foci are (0, √21) and (0, -√21).

The coordinates of the vertices are (0, 5) and (0, -5)

Length of major axis = 2a = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √21/5

Length of latus rectum = 2b²/a= (2×2²)/5 = (2×4)/5 = 8/5

Question - 3 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/16 + y²/9= 1

Answer - 3 : -

Given:

The equation is x²/16+ y²/9 = 1 or x²/4² + y²/3² = 1

Here, the denominator of x²/16 isgreater than the denominator of y²/9.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x²/a² +y²/b² =1, we get

a = 4 and b = 3.

c = √(a² – b²)

= √(16-9)

= √7

Then,

The coordinates of the foci are (√7, 0) and (-√7, 0).

The coordinates of the vertices are (4, 0) and (-4, 0)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = c/a = √7/4

Length of latus rectum = 2b²/a= (2×3²)/4 = (2×9)/4 = 18/4 = 9/2

Question - 4 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/25 + y²/100= 1

Answer - 4 : -

Given:

The equation is x²/25+ y²/100 = 1

Here, the denominator of y²/100 isgreater than the denominator of x²/25.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x²/b² +y²/a² =1, we get

b = 5 and a =10.

c = √(a² – b²)

= √(100-25)

= √75

= 5√3

Then,

The coordinates of the foci are (0, 5√3) and (0, -5√3).

The coordinates of the vertices are (0, √10) and (0, -√10)

Length of major axis = 2a = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

Eccentricity, e = c/a = 5√3/10 = √3/2

Length of latus rectum = 2b²/a= (2×5²)/10 = (2×25)/10 = 5

Question - 5 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/49 + y²/36= 1

Answer - 5 : -

Given:

The equation is x²/49+ y²/36 = 1

Here, the denominator of x²/49 isgreater than the denominator of y²/36.

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x²/a² +y²/b² =1, we get

b = 6 and a =7

c = √(a² – b²)

= √(49-36)

= √13

Then,

The coordinates of the foci are (√13, 0) and (-√3, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = c/a = √13/7

Length of latus rectum = 2b²/a= (2×6²)/7 = (2×36)/7 = 72/7

Question - 6 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
x²/100 + y²/400= 1

Answer - 6 : -

Given:

The equation is x²/100+ y²/400 = 1

Here, the denominator of y²/400 isgreater than the denominator of x²/100.

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x²/b² +y²/a² =1, we get

b = 10 and a =20.

c = √(a² – b²)

= √(400-100)

= √300

= 10√3

Then,

The coordinates of the foci are (0, 10√3) and (0, -10√3).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

Eccentricity, e = c/a = 10√3/20 = √3/2

Length of latus rectum = 2b²/a= (2×10²)/20 = (2×100)/20 = 10

Question - 7 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
36x² + 4y² =144

Answer - 7 : -

Given:

The equation is 36x² +4y² = 144 or x²/4 + y²/36= 1 or x²/2² +y²/6² =1

Here, the denominator of y²/6² is greater than the denominator of x²/2².

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x²/b² +y²/a² =1, we get

b = 2 and a = 6.

c = √(a² – b²)

= √(36-4)

= √32

= 4√2

Then,

The coordinates of the foci are (0, 4√2) and (0, -4√2).

The coordinates of the vertices are (0, 6) and (0, -6)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = 4√2/6 = 2√2/3

Length of latus rectum = 2b²/a= (2×2²)/6 = (2×4)/6 = 4/3

Question - 8 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
16x² + y² =16

Answer - 8 : -

Given:

The equation is 16x² +y² = 16 or x²/1 + y²/16= 1 or x²/1² +y²/4² =1

Here, the denominator of y²/4² is greater than the denominator of x²/1².

So, the major axis is along the y-axis, while the minor axis isalong the x-axis.

On comparing the given equation with x²/b² +y²/a² =1, we get

b =1 and a =4.

c = √(a² – b²)

= √(16-1)

= √15

Then,

The coordinates of the foci are (0, √15) and (0, -√15).

The coordinates of the vertices are (0, 4) and (0, -4)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/a = √15/4

Length of latus rectum = 2b²/a= (2×1²)/4 = 2/4 = ½

Question - 9 : - Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
4x² + 9y² =36

Answer - 9 : -

Given:

The equation is 4x² +9y² = 36 or x²/9 + y²/4= 1 or x²/3² +y²/2² =1

Here, the denominator of x²/3² is greater than the denominator of y²/2².

So, the major axis is along the x-axis, while the minor axis isalong the y-axis.

On comparing the given equation with x²/a² +y²/b² =1, we get

a =3 and b =2.

c = √(a² – b²)

= √(9-4)

= √5

Then,

The coordinates of the foci are (√5, 0) and (-√5, 0).

The coordinates of the vertices are (3, 0) and (-3, 0)

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

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/a = √5/3

Length of latus rectum = 2b²/a= (2×2²)/3 = (2×4)/3 = 8/3

Question - 10 : - Find the equation for the ellipse that satisfies the given conditions:
Vertices (± 5, 0), foci (± 4, 0)

Answer - 10 : -

Given:

Vertices (± 5, 0) and foci (± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the form x²/a² +y²/b² = 1, where ‘a’ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a² = b²+ c².

So, 5² = b²+ 4²

25 = b² + 16

b² = 25 – 16

b = √9

= 3

∴ Theequation of the ellipse is x²/5² + y²/3² =1 or x²/25 + y²/9 = 1

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