Class 11 Maths Ncert Solutions Chapter 11 Ex 11.3 Conic Sections PDF

Class 11 Maths Ncert Solutions Ex 11.3

Class 11 Maths Ncert Solutions Chapter 11 Ex 11.3

 Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x225+y29=1.

 

Sol:

Given:

x225+y29=1 . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x225 > the denominator of y29

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1 . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=259=16=4

We get:

m = 5, n = 3

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (4, 0) and (- 4, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 6

Length of the latus rectum = 2n2m = 2.325=185

Eccentricity, e = cm = 45

 

 

Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x29+y216=1.

 

Sol:

Given:

x29+y216=1 . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x29 < the denominator of y216

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1 . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=169=7

We get,:

m = 4, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 4), (0, – 4)

The foci’s coordinates are (0, c) and (0, – c) = (7, 0) and (-7, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 6

Length of the latus rectum = 2n2m = 2.324=92

Eccentricity, e = cm = 74

 

 

Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x225+y24=1.

 

Sol:

Given:

x225+y24=1 . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x225 > the denominator of y24

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1 . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=254=21

We get:

m = 5, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (5, 0), (- 5, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (21, 0) and (-21, 0)

Length of the axis:

Major axis = 2m = 10

Minor axis = 2n = 4

Length of the latus rectum = 2n2m=2.225=85

Eccentricity, e = cm = 215

 

 

Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x236+y2121=1.

 

Sol:

Given:
x236+y2121=1 . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x236 < the denominator of y2121

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1 . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=12136=85

We get:

m = 11, n = 6

The vertices coordinates are (0, m) and (0, – m) = (0, 11), (0, – 11)

The foci’s coordinates are (0, c) and (0, – c) = (0, 85) and (0, –85)

Length of the axis:

Major axis = 2m = 22

Minor axis = 2n = 12

Length of the latus rectum = 2n2m = 2.6211=7211

Eccentricity, e = cm = 8511

 

 

Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x264+y225=1.

 

Sol:

Given:

x264+y225=1 . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x264 > the denominator of y225

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1 . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=6425=39

We get:

m = 8, n = 5

The vertices coordinates are (m, 0) and (- m, 0) = (8, 0), (- 8, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (39, 0) and (-39, 0)

Length of the axis:

Major axis = 2m = 16

Minor axis = 2n = 10

Length of the latus rectum = 2n2m = 2.528=254

Eccentricity, e = cm = 398

 

 

Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse x2400+y2100=1.

 

Sol:

Given:

x2400+y2100=1 . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x2400 > the denominator of y2100

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1 . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=400100=300=103

We get:

m = 20, n = 10

The vertices coordinates are (m, 0) and (- m, 0) = (20, 0), (- 20, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (103, 0) and (-103, 0)

Length of the axis:

Major axis = 2m = 40

Minor axis = 2n = 20

Length of the latus rectum = 2n2m = 2.10220=10

Eccentricity, e = cm = 10320=32

 

 

Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x2 + 9y2 = 441

 

Sol:

Given:

49x2 + 9y2 = 441

x29+y249=1  . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x29 < the denominator of y249

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1 . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=499=40=210

We get:

m = 7, n = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 7), (0, – 7)

The foci’s coordinates are (0, c) and (0, – c) = (0, 210) and (0, 210)

Length of the axis:

Major axis = 2m = 14

Minor axis = 2n = 6

Length of the latus rectum = 2n2m = 2.327=187

Eccentricity, e = cm = 2107

 

 

Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + 16y2 = 64

 

Sol:

Given:

4x2 + 16y2 = 64

x216+y24=169 . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x216 > the denominator of y24

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

x2m2+y2n2=1 . . . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=164=12=23

We get:

m = 4, n = 2

The vertices coordinates are (m, 0) and (- m, 0) = (4, 0), (- 4, 0)

The foci’s coordinates are (c, 0) and (- c, 0) = (23, 0) and (-23, 0)

Length of the axis:

Major axis = 2m = 8

Minor axis = 2n = 4

Length of the latus rectum = 2n2m = 2.424=8

Eccentricity, e = cm = 234=32

 

 

Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x2 + y2 = 4

 

Sol:

Given:

4x2 + y2 = 4

x21+y24=1 . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.

As we know that the denominator of x21 < the denominator of y24

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

x2n2+y2m2=1 . . . . . . . . . . . . . (2)

Now, on comparing equation (1) and equation (2) we will get:

c=m2n2=41=3

We get:

m = 2, n = 1

The vertices coordinates are (0, m) and (0, – m) = (0, 2), (0, – 2)

The foci’s coordinates are (0, c) and (0, – c) = (0, 3) and (0, –3)

Length of the axis:

Major axis = 2m = 4

Minor axis = 2n = 2

Length of the latus rectum = 2n2m = 2.122=1

Eccentricity, e = cm = 32

 

 

Q.10: For the given condition obtain the equation of the ellipse.

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

 

Sol:

Given:

(i) Vertices (±6, 0)

(ii) Foci (±3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form x2m2+y2n2=1 . . . . . . . . . . . . . . . . . (1)

We get:

m = 6 (semi major axis), c = 3

As we know:

m2 = n2 + c2

62 = n2 + 32

n2 = 62 – 32

n2 = 36 – 9 = 27

n = 27=33

Hence, x236+y227=1  is the equation of the ellipse.

 

 

Q.11: For the given condition obtain the equation of the ellipse.

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

 

Sol:

Given:

(i) Vertices (0, ±8)

(ii) Foci ( 0, ±4)

The vertices are represented along y – axis.

The required equation of the ellipse is of the form  x2n2+y2m2=1 . . . . . . . . . . . . . . .(1)

We get:

m = 8 (semi major axis), c = 4

As we know:

m2 = n2 + c2

82 = n2 + 42

n2 = 82 – 42

n2 = 64 – 16 = 48

n = 48=43

Hence, x248+y264=1 is the equation of the ellipse.

 

 

Q.12: For the given condition obtain the equation of the ellipse.

(i) Vertices (±5, 0)

(ii) Foci (±3, 0)

 

Sol:

Given:

(i) Vertices (± 5, 0)

(ii) Foci (± 3, 0)

The vertices are represented along x – axis.

The required equation of the ellipse is of the form  x2m2+y2n2=1 . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), c = 3

As we know:

m2 = n2 + c2

52 = n2 + 32

n2 = 52 – 32

n2 = 25 – 9 = 16

n = 16=4

Hence, x225+y216=1 is the equation of the ellipse.

 

 

Q.13: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

 

Sol:

Given:

(i) Coordinates of major axis (±5, 0)

(ii) Coordinates of minor axis (0, ±3)

The major axis is represented along x – axis.

The required equation of the ellipse is of the form

x2m2+y2n2=1 . . . . . . . . . . . . . . . (1)

We get:

m = 5 (semi major axis), n = 3

Hence, x225+y216=1 is the equation of the ellipse.

 

 

Q.14: For the given condition obtain the equation of the ellipse.

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±3, 0)

 

Sol:

Given:

(i) Coordinates of major axis (0, ±2)

(ii) Coordinates of minor axis (±3, 0)

The major axis is represented along y – axis.

The required equation of the ellipse is of the form  x2n2+y2m2=1 . . . . . . . . . . . . . . . . . (1)

We get:

m = 2 (semi major axis), n = 3

Hence, x23+y24=1 is the equation of the ellipse.

 

 

Q.15: For the given condition obtain the equation of the ellipse.

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

 

Sol:

Given:

(i) Length of major axis 30

(ii) Coordinates of foci (±4, 0)

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form x2m2+y2n2=1 . . . . . . . . . . . . . . . . . . . (1)

We get:

2m = 30 (semi major axis),

m = 15

c = 4

As we know:

 m2 = n2 + c2

152 = n2 + 42

n2 = 152 – 42

n2 = 225 – 16 = 209

n = 209

Hence, x2225+y2209=1 is the equation of the ellipse.

 

 

Q.16: For the given condition obtain the equation of the ellipse.

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

 

Sol:

Given:

(i) Length of minor axis 26

(ii) Coordinates of foci (0, ±9)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form

x2n2+y2m2=1 . . . . . . . . . . . . . (1)

We get:

2n = 26 (semi major axis),

n = 13

c = 9

As we know:

m2 = n2 + c2

m2 = 132 + 92

m2 = 169 + 81

m2 = 250

m = 250 = 510

Hence, x2169+y2250=1 is the equation of the ellipse.

 

 

Q.17: For the given condition obtain the equation of the ellipse.

(i) Coordinates of foci (±4, 0)

(ii) m = 6

 

Sol:

Given:

(i) Coordinates of foci (±4, 0)

(ii) m = 6

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form

x2m2+y2n2=1 . . . . . . . . . . . . . . . (1)

We get:

c = 4

m = 6

As we know:

m2 = n2 + c2

62 = n2 + 42

n2 = 36 – 16

n2 = 20

n = 20 = 25

Hence, x236+y220=1 is the equation of the ellipse.

 

 

Q.18: For the given condition obtain the equation of the ellipse.

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0)

 

Sol:

Given:

(i) n = 2, c = 3 (on the x – axis)

(ii) Coordinates of centre (0, 0) i.e., centre is at origin.

The major axis are represented along x – axis as foci is along x – axis

The required equation of the ellipse is of the form:

x2m2+y2n2=1 . . . . . . . . . . . . . (1)

We have:

n = 2, c = 3

As we know:

m2 = n2 + c2

m2 = 22 + 32

m2 = 4 + 9

m2 = 13

m = 13

Hence, x213+y24=1 is the equation of the ellipse.

 

 

Q.19: For the given condition obtain the equation of the ellipse.

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

 

Sol:

Given:

(i) Centre is at origin and major axis is along the y – axis

(ii) It passes through the points (1, 6) and (3, 2)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

x2n2+y2m2=1 . . . . . . . . . . . . . . . (1)

As the ellipse passes through (2, 1) and (2, 3) points, then

12n2+62m2=1 32n2+22m2=1

1n2+36m2 = 1 . . . . . . . . . . . . . . . (a)

9n2+4m2 = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 40, n2 = 10

Hence, y240+x210=1 is the equation of the ellipse.

 

 

Q.20: For the given condition obtain the equation of the ellipse.

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

 

Sol:

Given:

(i) Major axis is represented along x – axis

(ii) It passes through the points (6, 2) and (4, 3)

The major axis are represented along y – axis as foci is along y – axis

The required equation of the ellipse is of the form:

x2m2+y2n2=1 . . . . . . . . . . . . . . . . (1)

As the ellipse passes through (6, 2) and (4, 3) points, then

62m2+22n2=1 42m2+32n2=1

36m2+4n2 = 1 . . . . . . . . . . . . . . . . (a)

16m2+9n2 = 1 . . . . . . . . . . . . . . . . (b)

Now, on comparing equation (a) and equation (b) we will get:

m2 = 52 and n2 = 13

Hence, y213+x252=1 is the equation of the ellipse.