Chapter 11 Conic Sections Ex-11.1 |
Chapter 11 Conic Sections Ex-11.2 |
Chapter 11 Conic Sections Ex-11.4 |

**x ^{2}/36 + y^{2}/16= 1**

**Answer
1** :

Given:

The equation is x^{2}/36+ y^{2}/16 = 1

Here, the denominator of x^{2}/36 isgreater than the denominator of y^{2}/16.

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

On comparing the given equation with x^{2}/a^{2} +y^{2}/b^{2} =1, we get

a = 6 and b = 4.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×16)/6 = 16/3

**x ^{2}/4 + y^{2}/25= 1**

**Answer
2** :

Given:

The equation is x^{2}/4 +y^{2}/25 = 1

Here, the denominator of y^{2}/25 isgreater than the denominator of x^{2}/4.

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

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 5 and b = 2.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×2^{2})/5 = (2×4)/5 = 8/5

**x ^{2}/16 + y^{2}/9= 1**

**Answer
3** :

Given:

The equation is x^{2}/16+ y^{2}/9 = 1 or x^{2}/4^{2} + y^{2}/3^{2} = 1

Here, the denominator of x^{2}/16 isgreater than the denominator of y^{2}/9.

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

On comparing the given equation with x^{2}/a^{2} +y^{2}/b^{2} =1, we get

a = 4 and b = 3.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×3^{2})/4 = (2×9)/4 = 18/4 = 9/2

**x ^{2}/25 + y^{2}/100= 1**

**Answer
4** :

Given:

The equation is x^{2}/25+ y^{2}/100 = 1

Here, the denominator of y^{2}/100 isgreater than the denominator of x^{2}/25.

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

On comparing the given equation with x^{2}/b^{2} +y^{2}/a^{2} =1, we get

b = 5 and a =10.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×5^{2})/10 = (2×25)/10 = 5

**x ^{2}/49 + y^{2}/36= 1**

**Answer
5** :

Given:

The equation is x^{2}/49+ y^{2}/36 = 1

Here, the denominator of x^{2}/49 isgreater than the denominator of y^{2}/36.

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

On comparing the given equation with x^{2}/a^{2} +y^{2}/b^{2} =1, we get

b = 6 and a =7

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×6^{2})/7 = (2×36)/7 = 72/7

**x ^{2}/100 + y^{2}/400= 1**

**Answer
6** :

Given:

The equation is x^{2}/100+ y^{2}/400 = 1

Here, the denominator of y^{2}/400 isgreater than the denominator of x^{2}/100.

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

On comparing the given equation with x^{2}/b^{2} +y^{2}/a^{2} =1, we get

b = 10 and a =20.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×10^{2})/20 = (2×100)/20 = 10

**36x ^{2} + 4y^{2} =144**

**Answer
7** :

Given:

The equation is 36x^{2} +4y^{2} = 144 or x^{2}/4 + y^{2}/36= 1 or x^{2}/2^{2} +y^{2}/6^{2} =1

Here, the denominator of y^{2}/6^{2} is greater than the denominator of x^{2}/2^{2}.

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

On comparing the given equation with x^{2}/b^{2} +y^{2}/a^{2} =1, we get

b = 2 and a = 6.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×2^{2})/6 = (2×4)/6 = 4/3

**16x ^{2} + y^{2} =16**

**Answer
8** :

Given:

The equation is 16x^{2} +y^{2} = 16 or x^{2}/1 + y^{2}/16= 1 or x^{2}/1^{2} +y^{2}/4^{2} =1

Here, the denominator of y^{2}/4^{2} is greater than the denominator of x^{2}/1^{2}.

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

On comparing the given equation with x^{2}/b^{2} +y^{2}/a^{2} =1, we get

b =1 and a =4.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×1^{2})/4 = 2/4 = ½

**4x ^{2} + 9y^{2} =36**

**Answer
9** :

Given:

The equation is 4x^{2} +9y^{2} = 36 or x^{2}/9 + y^{2}/4= 1 or x^{2}/3^{2} +y^{2}/2^{2} =1

Here, the denominator of x^{2}/3^{2} is greater than the denominator of y^{2}/2^{2}.

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

On comparing the given equation with x^{2}/a^{2} +y^{2}/b^{2} =1, we get

a =3 and b =2.

c = √(a^{2} – b^{2})

= √(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 = 2b^{2}/a= (2×2^{2})/3 = (2×4)/3 = 8/3

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 x^{2}/a^{2} +y^{2}/b^{2} = 1, where ‘a’ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a^{2} = b^{2 }+ c^{2}.

So, 5^{2} = b^{2 }+ 4^{2}

25 = b^{2} + 16

b^{2} = 25 – 16

b = √9

= 3

∴ Theequation of the ellipse is x^{2}/5^{2} + y^{2}/3^{2} =1 or x^{2}/25 + y^{2}/9 = 1

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