**Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola x225–y216=1.**

** **

**Sol:**

**Given:**

**. . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . . . . . .(2)**

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

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 =**

**Eccentricity, e =**

**Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola x21–y28=1.**

** **

**Sol:**

**Given:**

**. . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . . . . . . . . . . .(2)**

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

**m =** 1and **n =**

As we know:

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 =**

**Eccentricity, e =**

**Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y ^{2} – 4 x^{2} = 100**

** **

**Sol:**

**Given:**

**25 y ^{2} – 4 x^{2} = 100**

**. . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is

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

**m = 2** and **n = 5.**

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,

**Length of the latus rectum =**

**Eccentricity, e =**

**Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x ^{2} – 4 y^{2} = 36**

** **

**Sol:**

**Given:**

**9 x ^{2} – 4 y^{2} = 36**

** . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . . . . (2)**

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

**m = 2** and **n = 3.**

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

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

**Length of the latus rectum =**

**Eccentricity, e =**

**Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y ^{2} – 16 x^{2} = 96**

** **

**Sol:**

**Given:**

**6 y ^{2} – 16 x^{2} = 96**

** . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . . . . . .(2)**

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

**m =** 4 and **n = **

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 =**

**Eccentricity, e =**

**Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y ^{2} – 9 x^{2} = 96**

** **

**Sol:**

**Given:**

**6 y ^{2} – 16 x^{2} = 96**

** . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The specific equation for hyperbola is

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

**m =** 4 and **n =**

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 =**

**Eccentricity, e =**

**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 **. . . . . . . . . . . . . . . . (1)**

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

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

We know that:

**c ^{2} = m^{2} + n^{2}**

4^{2} = 3^{2} + n^{2}

n^{2} = 16 – 9 = 5

**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 **. . . . . . . . . . . . . . . . . . . (1)**

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

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

We know that:

**c ^{2} = m^{2} + n^{2}**

9^{2} = 7^{2} + n^{2}

**n ^{2} =** 81 – 49

**= 32**

**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 **. . . . . . . . . . . . . . . . . . . . . . (1)**

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

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

We know that:

**c ^{2} = m^{2} + n^{2}**

11^{2} = 8^{2} + n^{2}

**n ^{2} =** 121 – 64

**= 57**

**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 **. . . . . . . . . . . . . . . . . . (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:

**c ^{2} = m^{2} + n^{2}**

7^{2} = 6^{2} + n^{2}

**n ^{2} =** 49 – 36

**= 13**

**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 **. . . . . . . . . . . . . . . . (1)**

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

As, the **conjugate axis is of length 16,**

2 m = 16

**m = 8**

**Therefore, m ^{2} = 64**

We know that:

**c ^{2} = m^{2} + n^{2}**

9^{2} = 8^{2} + n^{2}

**n ^{2} **= 81 – 64

**= 17**

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

**(i) Coordinates of the foci (****±**

**(ii) The length of the latus rectum is 8**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (****±**

**(ii) The length of the latus rectum is 8**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . (1)**

And, the **coordinates of the foci** (±**c = 23–√**

As, the **latus rectum is of length 8**

We know that:

**m ^{2} + n^{2} = c^{2}**

m^{2} + 4m =

m^{2} + 4m = 12

m^{2} + 4m – 12 = 0

m^{2 }+ 6m – 2m – 12 = 0

m (m + 6) – 2 (m + 6) = 0

m = – 6 and m = 2

**m is non – negative** so, **m = 2** and **n ^{2} = 8 [Since, n^{2} = 12 – 2^{2} = 8] **

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

**(i) Coordinates of the foci (****±**

**(ii) The length of the latus rectum is 10**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (****±**

**(ii) The length of the latus rectum is 10**

The **specific equation for hyperbola** is **. . . . . . . . . . . . . . . . . (1)**

And, coordinates of the foci (±**c = 26–√**

As, the **latus rectum is of length 10**

Therefore,

We know that:

** m ^{2} + n^{2} = c^{2}**

m^{2} + 5 m =

m^{2} + 5 m = 24

m^{2} + 5 m – 24 = 0

m^{2 }+ 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 **n ^{2} =** 5m

**= 15**

** **

**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 **. . . . . . . . . . . . . . . . . (1)**

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

As, **the eccentricity (e) =**

**c =**

**We know that:**

** m ^{2} + n^{2} = c^{2}**

9^{2} + n^{2} =

n^{2} = ^{2}

n^{2 }=

n^{2 }=

** **

**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, ****±**

**(ii) It passes through (2, 3)**

The **specific** **equation for hyperbola** is **. . . . . . . . . . . . (1)**

And,** coordinates of the foci **(0, ±

We know that:

** m ^{2} + n^{2} = c^{2}**

n^{2} = c^{2} – m^{2}

n^{2} = 10 – m^{2} **. . . . . . . . . . . . (2)**

**As the hyperbola passes through (2, 3)**

**. . . . . . . . . . . (3)**

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

9 (10 – m^{2}) – 4m^{2} = m^{2} (10 – m^{2})

m^{4} – 23m^{2} + 90 = 0

m^{4} – 18m^{2} – 5m^{2} + 90 = 0

(m^{2} – 5) (m^{2} – 18) = 0

**m ^{2 }= 5 or 18**

As in hyperbola:

m^{2} < c^{2}

**m ^{2 }= 5**

**n ^{2 }=** 10 – 5 [From Equation (2)]

**= 5**