Class 11 Maths Ncert Solutions Chapter 11 Ex 11.4 Conic Sections PDF

Class 11 Maths Ncert Solutions Ex 11.4

Class 11 Maths Ncert Solutions Chapter 11 Ex 11.4

Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola x225y216=1.

 

Sol:

Given:

x225y216=1 . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . . .(2)

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

c=m2+n2=25+16=41

We get:

m = 5 and n = 4

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

The foci’s coordinates are (c, 0) and (- c, 0) = (41, 0) and (-41, 0)

Length of the latus rectum = 2n2m = 2.425=325

Eccentricity, e = cm = 415

 

 

Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola x21y28=1.

 

Sol:

Given:

x21y28=1 . . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . . . . . . . .(2)

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

m = 1and  n = 8

As we know:

c=m2+n2=1+8=9 = 3

The vertices coordinates are (0, m) and (0, – m) = (0, 1), (0, – 1)

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

Length of the latus rectum = 2n2m = 161=16

Eccentricity, e = cm = 31 = 3

 

 

Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y2 – 4 x2 = 100

 

Sol:

Given:

25 y2 – 4 x2 = 100

y24x225=1 . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . .(2)

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

m = 2 and n = 5.

c=m2+n2=4+25=29

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

The foci’s coordinates are (0, c) and (0, – c) = (0, 29) and (0, –29)

Length of the latus rectum = 2.n2m = 2.522=502 = 25

Eccentricity, e = cm = 292

 

 

Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x2 – 4 y2 = 36

 

Sol:

Given:

9 x2 – 4 y2 = 36

x24y29=1 . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . (2)

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

m = 2 and n = 3.

c=m2+n2=4+3=13

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

The foci’s coordinates are (c, 0) and (- c, 0) = (13, 0) and (-13, 0)

Length of the latus rectum = 2n2m = 2.322=9

Eccentricity, e = cm = 132

 

 

Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y2 – 16 x2 = 96

 

Sol:

Given:

6 y2 – 16 x2 = 96

y216x26=1 . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . . . . .(2)

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

m = 4 and n = 6

c=m2+n2=4+6=10

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

The foci’s coordinates are (0, c) and (0, – c) = (0, 10) and (0, –10)

Length of the latus rectum = 2.n2m = 2.624=124 = 3

Eccentricity, e = cm = 104

 

 

Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y2 – 9 x2 = 96

 

Sol:

Given:

6 y2 – 16 x2 = 96

y216x26=1 . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . . . . . . . (2)

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

m = 4 and n = 6

c=m2+n2=4+6=10

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

The foci’s coordinates are (0, c) and (0, – c) = (0, 10) and (0, –10)

Length of the latus rectum = 2.n2m = 2.624=124 = 3

Eccentricity, e = cm = 104

 

 

Q.7: For the given condition obtain the equation of the hyperbola.

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

 

Sol:

Given:

(i) Coordinates of the vertices (±3, 0)

(ii) Coordinates of the foci (±4, 0)

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (±3, 0), m = 3

And, coordinates of the foci (±4, 0), c = 4

We know that:

c2 = m2 + n2

42 = 32 + n2

n2 = 16 – 9 = 5

x29y25=1 is the equation of the hyperbola.

 

 

Q.8: For the given condition obtain the equation of the hyperbola.

 (i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

 

Sol:

Given:

(i) Coordinates of the vertices (0, ±7)

(ii) Coordinates of the foci (0, ±9)

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±7), m = 7

And, coordinates of the foci (0, ±9), c = 9

We know that:

c2 = m2 + n2

92 = 72 + n2

n2 = 81 – 49 = 32

y249x232=1 is the equation of the hyperbola.

 

 

Q.9: For the given condition obtain the equation of the hyperbola.

 (i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

 

Sol:

Given:

(i) Coordinates of the vertices (0, ±8)

(ii) Coordinates of the foci (0, ±11)

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . . . . . . . . . (1)

As, coordinates of the vertices (0, ±8), m = 8

And, coordinates of the foci (0, ±11), c = 11

We know that:

c2 = m2 + n2

112 = 82 + n2

n2 = 121 – 64 = 57

y264x257=1 is the equation of the hyperbola.

 

 

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

 (i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

 

Sol:

Given:

(i) Coordinates of the foci (±7, 0),

(ii) The length of the transverse axis is 12

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±7, 0), c = 7

As, the transverse axis is of length 12,

2 m = 12

m = 6

We know that:

c2 = m2 + n2

72 = 62 + n2

n2 = 49 – 36 = 13

x236y213=1 is the equation of the hyperbola.

 

 

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

(i) Coordinates of the foci (0, ±8)

(ii) The length of the conjugate axis is 16

 

Sol:

Given:

(i) Coordinates of the foci (0, ±9)

(ii) The length of the conjugate axis is 16

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±9), c = 9

As, the conjugate axis is of length 16,

2 m = 16

m = 8

Therefore, m2 = 64

We know that:

c2 = m2 + n2

92 = 82 + n2

n2 = 81 – 64 = 17

y264x217=1 is the equation of the hyperbola.

 

 

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

(i) Coordinates of the foci (±23, 0 )

(ii) The length of the latus rectum is 8

 

Sol:

Given:

(i) Coordinates of the foci (±23, 0 )

(ii) The length of the latus rectum is 8

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . (1)

And, the coordinates of the foci23, 0 ), c = 23

As, the latus rectum is of length 8

2n2m=82n2=8mn2=4m

We know that:

m2 + n2 = c2

m2 + 4m = (23)2

m2 + 4m = 12

m2 + 4m – 12 = 0

m2 + 6m – 2m – 12 = 0

m (m + 6) – 2 (m + 6) = 0

m = – 6 and m = 2

m is non – negative so, m = 2 and n2 = 8 [Since, n2 = 12 – 22 = 8]

x24y28=1 is the equation of the hyperbola.

 

 

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

(i) Coordinates of the foci (±26, 0)

(ii) The length of the latus rectum is 10

 

Sol:

Given:

(i) Coordinates of the foci (±23, 0 )

(ii) The length of the latus rectum is 10

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±26, 0 ), c = 26

As, the latus rectum is of length 10

Therefore, 2n2m=102n2=10mn2=5m

We know that:

 m2 + n2 = c2

m2 + 5 m = (26)2

m2 + 5 m = 24

m2 + 5 m – 24 = 0

m2 + 8 m – 3 m – 24 = 0

m (m + 8) – 3 (m + 8) = 0

m = – 8 and m = 3

m is non – negative so, m = 3 and n2 = 5m = 15

x29y215=1 is the equation of the hyperbola.

 

 

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

(i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = 94

 

Sol:

Given:

 (i) Coordinates of the vertices (±9, 0)

(ii) Eccentricity (e) = 94

The specific equation for hyperbola is x2m2y2n2=1 . . . . . . . . . . . . . . . . . (1)

And, Coordinates of the vertices (±9, 0), m = 9

As, the eccentricity (e) = 94,

cm=94 c9=94

c = 814

We know that:

 m2 + n2 = c2

92 + n2 = (814)2

n2 = (814)2 – 92

n2 = 65611681

n2 = 6561129616=526516

x28116y25265=1 is the equation of the hyperbola.

 

 

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

(i) Coordinates of the foci (0, ± 10)

(ii) It passes through (2, 3)

 

Sol:

Given:

(i) Coordinates of the foci (0, ±10)

(ii) It passes through (2, 3)

The specific equation for hyperbola is y2m2x2n2=1 . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±10), c = 10

We know that:

 m2 + n2 = c2

n2 = c2 – m2

n2 = 10 – m2 . . . . . . . . . . . . (2)

As the hyperbola passes through (2, 3)

32m222n2=1

9m24n2=1 . . . . . . . . . . . (3)

Putting Equation (2) in equation (3), we get:

9m2410m2=1

9 (10 – m2) – 4m2 = m2 (10 – m2)

m4 – 23m2 + 90 = 0

m4 – 18m2 – 5m2 + 90 = 0

(m2 – 5) (m2 – 18) = 0

m2 = 5 or 18

As in hyperbola:

m2 < c2

m2 = 5

n2 = 10 – 5 [From Equation (2)] = 5

y25x25=1 is the equation of the hyperbola.