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$\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$.

Sol:

Given:

x225y216=1$\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1$ . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . .(2)

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

c=m2+n2=25+16=41$c = \sqrt{m^{2} + n^{2}} = \sqrt{25 + 16} = \sqrt{41}$

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$\sqrt{41}$, 0) and (-41$\sqrt{41}$, 0)

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.425=325$\frac{2.4^{2}}{5} = \frac{32}{5}$

Eccentricity, e = cm$\frac{c}{m}$ = 415$\frac{\sqrt{41}}{5}$

Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola x21y28=1$\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$.

Sol:

Given:

x21y28=1$\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1$ . . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . . . .(2)

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

m = 1and  n = 8$\sqrt{8}$

As we know:

c=m2+n2=1+8=9$c = \sqrt{m^{2} + n^{2}} = \sqrt{1 + 8} = \sqrt{9}$ = 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$\frac{2n^{2}}{m}$ = 161=16$\frac{16}{1} = 16$

Eccentricity, e = cm$\frac{c}{m}$ = 31$\frac{\sqrt{3}}{1}$ = 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$\frac{y^{2}}{4} – \frac{x^{2}}{25} = 1$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . .(2)

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

m = 2 and n = 5.

c=m2+n2=4+25=29$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 25} = \sqrt{29}$

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$\sqrt{29}$) and (0, –29$\sqrt{29}$)

Length of the latus rectum = 2.n2m$\frac{2.n^{2}}{m}$ = 2.522=502$\frac{2.5^{2}}{2} = \frac{50}{2}$ = 25

Eccentricity, e = cm$\frac{c}{m}$ = 292$\frac{\sqrt{29}}{2}$

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$\frac{x^{2}}{4} – \frac{y^{2}}{9} = 1$ . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (2)

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

m = 2 and n = 3.

c=m2+n2=4+3=13$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 3} = \sqrt{13}$

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

The foci’s coordinates are (c, 0) and (- c, 0) = (13$\sqrt{13}$, 0) and (-13$\sqrt{13}$, 0)

Length of the latus rectum = 2n2m$\frac{2n^{2}}{m}$ = 2.322=9$\frac{2.3^{2}}{2} = 9$

Eccentricity, e = cm$\frac{c}{m}$ = 132$\frac{\sqrt{13}}{2}$

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$\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$ . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . .(2)

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

m = 4 and n = 6$\sqrt{6}$

c=m2+n2=4+6=10$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$

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$\sqrt{10}$) and (0, –10$\sqrt{10}$)

Length of the latus rectum = 2.n2m$\frac{2.n^{2}}{m}$ = 2.624=124$\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$ = 3

Eccentricity, e = cm$\frac{c}{m}$ = 104$\frac{\sqrt{10}}{4}$

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$\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1$ . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.

The specific equation for hyperbola is y2m2x2n2=1$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . (2)

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

m = 4 and n = 6$\sqrt{6}$

c=m2+n2=4+6=10$c = \sqrt{m^{2} + n^{2}} = \sqrt{4 + 6} = \sqrt{10}$

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$\sqrt{10}$) and (0, –10$\sqrt{10}$)

Length of the latus rectum = 2.n2m$\frac{2.n^{2}}{m}$ = 2.624=124$\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}$ = 3

Eccentricity, e = cm$\frac{c}{m}$ = 104$\frac{\sqrt{10}}{4}$

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$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (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$\frac{x^{2}}{9} – \frac{y^{2}}{5} = 1$ 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$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . (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$\frac{y^{2}}{49} – \frac{x^{2}}{32} = 1$ 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$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . . . . . (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$\frac{y^{2}}{64} – \frac{x^{2}}{57} = 1$ 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$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . . (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$\frac{x^{2}}{36} – \frac{y^{2}}{13} = 1$ 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$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . (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$\frac{y^{2}}{64} – \frac{x^{2}}{17} = 1$ is the equation of the hyperbola.

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

(i) Coordinates of the foci (±23$2\sqrt{3}$, 0 )

(ii) The length of the latus rectum is 8

Sol:

Given:

(i) Coordinates of the foci (±23$2\sqrt{3}$, 0 )

(ii) The length of the latus rectum is 8

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . (1)

And, the coordinates of the foci23$2\sqrt{3}$, 0 ), c = 23$2\sqrt{3}$

As, the latus rectum is of length 8

2n2m=82n2=8mn2=4m$\frac{2n^{2}}{m} = 8 \\ 2n^{2} = 8 m \\ n^{2} = 4 m$

We know that:

m2 + n2 = c2

m2 + 4m = (23)2$(2\sqrt{3})^{2}$

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$\frac{x^{2}}{4} – \frac{y^{2}}{8} = 1$ is the equation of the hyperbola.

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

(i) Coordinates of the foci (±26$2\sqrt{6}$, 0)

(ii) The length of the latus rectum is 10

Sol:

Given:

(i) Coordinates of the foci (±23$2\sqrt{3}$, 0 )

(ii) The length of the latus rectum is 10

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1)

And, coordinates of the foci (±26$2\sqrt{6}$, 0 ), c = 26$2\sqrt{6}$

As, the latus rectum is of length 10

Therefore, 2n2m=102n2=10mn2=5m$\frac{2n^{2}}{m} = 10 \\ 2n^{2} = 10 m \\ n^{2} = 5m$

We know that:

m2 + n2 = c2

m2 + 5 m = (26)2$(2\sqrt{6})^{2}$

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$\frac{x^{2}}{9} – \frac{y^{2}}{15} = 1$ 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$\frac{9}{4}$

Sol:

Given:

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

(ii) Eccentricity (e) = 94$\frac{9}{4}$

The specific equation for hyperbola is x2m2y2n2=1$\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1$ . . . . . . . . . . . . . . . . . (1)

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

As, the eccentricity (e) = 94$\frac{9}{4}$,

cm=94$\frac{c}{m} = \frac{9}{4}$ c9=94$\frac{c}{9} = \frac{9}{4}$

c = 814$\frac{81}{4}$

We know that:

m2 + n2 = c2

92 + n2 = (814)2$(\frac{81}{4})^{2}$

n2 = (814)2$(\frac{81}{4})^{2}$ – 92

n2 = 65611681$\frac{6561}{16} – 81$

n2 = 6561129616=526516$\frac{6561 – 1296}{16} = \frac{5265}{16}$

x28116y25265=1$\frac{x^{2}}{81} – \frac{16 y^{2}}{5265} = 1$ 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$\sqrt{10}$)

(ii) It passes through (2, 3)

Sol:

Given:

(i) Coordinates of the foci (0, ±10$\sqrt{10}$)

(ii) It passes through (2, 3)

The specific equation for hyperbola is y2m2x2n2=1$\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1$ . . . . . . . . . . . . (1)

And, coordinates of the foci (0, ±10$\sqrt{10}$), c = 10$\sqrt{10}$

We know that:

m2 + n2 = c2

n2 = c2 – m2

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

As the hyperbola passes through (2, 3)

32m222n2=1$\frac{3^{2}}{m^{2}} – \frac{2^{2}}{n^{2}} = 1$

9m24n2=1$\frac{9}{m^{2}} – \frac{4}{n^{2}} = 1$ . . . . . . . . . . . (3)

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

9m2410m2=1$\frac{9}{m^{2}}-\frac{4}{10 – m^{2}}= 1$

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$\frac{y^{2}}{5} – \frac{x^{2}}{5} = 1$ is the equation of the hyperbola.