## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.4

Chapter 11 Conic Sections Exercise 11.4 Class 11 Maths NCERT Solutions is very helpful guide in boosting your marks in examinations. NCERT Solutions for Class 11 Maths will help in covering the important concepts related to syllabus through which you can solve difficult questions easily.

1. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola x

Comparing the equation x

we have, a = 4, b = 3

and c =Â âˆš(a

Therefore, the coordinates of the foci are (Â± 5, 0) and that of vertices are (Â± 4, 0).

Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

2. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

y

Comparing the equation y

we have, a = 3, b = 3âˆš3

and c =Â âˆš(a

Therefore, the coordinates of the foci are (0, Â± 6) and that of vertices are (0, Â± 3). Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

3. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola.

9y^{2}/16 â€“ y^{2}/9 = 1.**Answer**Comparing the equation x

^{2}/16 â€“ y^{2}/9 = 1 with the standard equation x^{2}/a^{2}Â â€“ y^{2}/b^{2}Â = 1,we have, a = 4, b = 3

and c =Â âˆš(a

^{2}Â + b^{2}) =Â âˆš(16 + 9) =Â âˆš25 = 5Therefore, the coordinates of the foci are (Â± 5, 0) and that of vertices are (Â± 4, 0).

Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

^{2}/a = 2(3)^{2}/4 = 9/2 units.2. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola

y

^{2}/9 â€“ x^{2}/27 = 1**Answer**Comparing the equation y

^{2}/9 â€“ x^{2}/27 = 1 with the standard equation.we have, a = 3, b = 3âˆš3

and c =Â âˆš(a

^{2}Â + b^{2}) =Â âˆš(9 + 27) =Â âˆš36 =Â âˆš6Therefore, the coordinates of the foci are (0, Â± 6) and that of vertices are (0, Â± 3). Also, the eccentricity e = c/a = 5/4 and the length of latus rectum = 2b

^{2}/a = 2(3âˆš3)^{2}/3 = 54/3 = 18 units.3. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola.

^{Â 2}Â â€“ 4x

^{Â 2}Â = 36

**Answer**

9y

^{Â 2}Â â€“ 4x

^{Â 2}Â = 36 is the equation of hyperbola

i.e., y

^{2}/4 â€“ x

^{2}/9 = 1

âˆ´Â a

^{Â 2Â }= 4, b

^{Â 2Â }= 9, âˆ´ c

^{Â 2Â }= a

^{Â 2Â }+ b

^{Â 2Â }= 4 + 9 = 13,

a = 2, b = 3, c =Â âˆš3

Axis is y-axis

Foci (0, Â±Â âˆš13), vertices = (0, Â± 2)

Eccentricity = e = c/a =Â âˆš13/2 ,

Latus rectum = 2b

^{2}/a = (2Â Ã—Â 9)/2 = 9

4. Find the coordinates of the foci and the vertices, the eccentricity and length of the latus rectum of the hyperbola

16x

^{Â 2Â }â€“ 9y

^{2}Â = 576

**Answer**

Equation of hyperbola is

16x

^{Â 2}Â â€“ 9y

^{Â 2}Â = 576

or x

^{2}/36 â€“ y

^{2}/64 = 1

âˆ´Â a

^{Â 2Â }= 36, b

^{Â 2Â }= 64, c

^{Â 2Â }= 36 + 64 = 100

âˆ´Â a = 6, b = 8, c = 10

Axis is x-axis

Foci are (Â± 10, 0)

Vertices are (Â± 6, 0)

Eccentricity = c/a = 10/6 = 5/3

Latus rectum = 2b

^{2}/a = (2Â Ã—Â 64)/6 = 64/3

5. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas

5y

^{Â 2Â }â€“ 9x

^{Â 2}Â = 36

**Answer**

Equation of hyperbolas is 5y

^{Â 2Â }â€“ 9x

^{Â 2}Â = 36

=> y

^{2}/36/5 - x

^{2}/4 = 1 => -x

^{2}/4 + y

^{2}/36/5 = 1

âˆ´Â b

^{2}Â = 36/5 , a

^{Â 2}Â = 4, c

^{Â 2}Â = b

^{Â 2}Â + a

^{Â 2}

= 4 + 36/5 = 56/5

âˆ´Â b = 6/âˆš5 , a = 2, c = 2âˆš14/âˆš5

Axis is along y-axisÂ âˆ´Foci are (0, Â± 2âˆš14/âˆš5) ,

vertices are (0, Â± 6/âˆš5)

Eccentricity, e = c/b = 2âˆš14/âˆš5Â Ã—Â âˆš5/6 =Â âˆš14/3;

Latus rectum = 2a

^{2}/b = 2Â Ã—Â 4/6/âˆš5 = 4âˆš5/3

6. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola

49y

^{Â 2}Â â€“ 16x

^{Â 2}Â = 784

**Answer**

Dividing the equation by 784 on both sides, we have

49y

^{2}/a â€“ 16x

^{2}/784 = 1 => y

^{2}/16 â€“ x

^{2}/49 = 1

Comparing the equation with the standard equation y

^{2}/a

^{2}Â â€“ x

^{2}/b

^{2}Â = 1 we find that a = 4, b = 7 and

c =Â âˆš(a

^{2Â }+ b

^{2Â }) =Â âˆš( 16 + 49) =Â âˆš65

Therefore the coordinates of the foci are (0, Â±Â âˆš65) and that of vertices are (0, Â± 4). Also the eccentricity e = c/a =Â âˆš65/4 and the length of latus rectum is 2b

^{2}/a = 2(7)

^{2}/4 = 49/2 units.

7. Find the equation of the hyperbola satisfying the given conditions:

Vertices (Â± 2, 0), foci (Â± 3, 0)

**Answer**

Since the foci are on x-axis, the equation of the hyperbola is of the form

x

^{2}/a

^{2}Â â€“ y

^{2}/b

^{2}Â = 1

Given: vertices are (Â± 2, 0), a = 2

Also, since foci are (Â± 3, 0), c = 3 and

bÂ

^{2}Â = c

^{Â 2Â }â€“ aÂ

^{2}Â = 9 â€“ 4 = 5

Therefore, the equation of the hyperbola is

x

^{2}/4 â€“ y

^{2}/5 = 1

8. Find the equation of the hyperbola satisfying the given conditions: Vertices (0, Â± 5), foci (0, Â± 8)

**Answer**

Since the foci are on y-axis, the equation of the hyperbola is of the form

y

^{2}/a

^{2}Â â€“ x

^{2}/b

^{2}Â = 1

Given: vertices are (0, Â± 5), a = 5

Also, since foci are (0, Â± 8), c = 8 and

b

^{Â 2Â }= c

^{Â 2Â }â€“ a

^{Â 2Â }= 64 â€“ 25 = 39

Therefore, the equation of the hyperbola is

x

^{2}/25 â€“ y

^{2}/39 = 1

9. Find the equation of the hyperbola satisfying the given conditions:

Vertices (0, Â± 3), foci (0, Â± 5)

**Answer**

Vertices are (0, Â± 3) âˆ´Â a = 3

Foci are (0, Â±5)

âˆ´Â c = 5, b

^{Â 2Â }= c

^{Â 2Â }â€“ a

^{Â 2Â }= 25 â€“ 9 = 16

âˆ´Â b = 4

âˆ´Â Equation of hyperbola is (axis being y-axis)

y

^{2}/9 â€“ x

^{2}/16 = 1

10. Find the equation of the hyperbola satisfying the given conditions:

Foci (Â± 5, 0), the transverse axis is of length 8.

**Answer**

Foci are (Â± 5, 0) âˆ´Â c = 5

Transverse axis = 8 âˆ´Â a = 4

b

^{Â 2Â }= c

^{Â 2}Â â€“ a

^{Â 2Â }= 25 â€“ 16 = 9Â âˆ´Â b = 3

Axis of hyperbola is x-axis since (Â± 5, 0) lies on it.

x

^{2}/16 â€“ y

^{2}/9 = 1

11. Find the equation of the hyperbola satisfying the given conditions:

Foci (0, Â± 13), the conjugate axis is of length 24.

**Answer**

Since the foci are on y-axis, the equation of the hyperbola is of the form

y

^{2}/a

^{2}Â â€“ x

^{2}/b

^{2}Â = 1

Given: foci are (0, Â± 13), c = 13

Length of conjugate axis = 2b = 24 => b = 12

or b

^{Â 2}Â = c

^{Â 2}Â â€“ a

^{Â 2}Â or 144 = 169 â€“ a

^{Â 2}

or a

^{Â 2}Â = 25 or a = 5

The equation of the hyperbola is

y

^{2}/25 â€“ x

^{2}/144 = 1

or 144 y

^{Â 2}Â â€“ 25x

^{Â 2}Â = 3600

12. Find the equation of the hyperbola satisfying the given conditions:

Foci (Â± 3âˆš5, 0), the latus rectum is of length 8.

**Answer**

Since the foci are on x-axis, the equation of the hyperbola is of the form

x

^{2}/a

^{2}Â â€“ y

^{2}/b

^{2}Â = 1

Given: foci are (Â± 3âˆš5, 0), c = 3âˆš5

and length of latus rectum = 2b

^{2}/a = 8

As b

^{Â 2}Â = 4a

We have c

^{Â 2}Â = a

^{Â 2}Â + b

^{Â 2}

or 45 = a

^{Â 2}Â + 4 a

or a

^{Â 2}Â + 4 a â€“ 45 = 0

or a

^{Â 2}Â + 9a â€“ 5a â€“ 45 = 0

or a(a + 9) â€“ 5 (a â€“ 9) = 0

or (a + 9)(a â€“ 5) = 0

or a = â€“9 or a = 5

Since a cannot be negative, we take a = 5 and so b

^{Â 2}Â = 20

Therefore, the equation of the required hyperbola is x

^{2}/25 â€“ y

^{2}/20 = 1 => 4x

^{2}Â â€“ 5y

^{2}Â = 100

13. Find the equation of hyperbola satisfying the given condition:

Foci (Â± 4, 0), the latus rectum is of length 12.

**Answer**

Foci are (Â± 4, 0) âˆ´Â c = 4

or c

^{2}Â = a

^{Â 2}Â + b

^{Â 2}âˆ´Â 16 = a

^{Â 2Â }+ b

^{Â 2Â }â€¦ (i)

Latus rectum = 2b

^{2}/a = 12

âˆ´Â b

^{Â 2Â }= 6a â€¦ (ii)

Eliminating b

^{Â 2Â }from (i) and (ii)

âˆ´Â 16 = a

^{Â 2Â }+ 6a or a

^{Â 2Â }+ 6a â€“ 16 = 0

or (a + 8) (a â€“ 2) = 0

a

^{2}â€“ 8, âˆ´Â a = 2 âˆ´Â b

^{Â 2Â }= 6a = 6 Ã— 2 = 12

a

^{Â 2Â }= 4, b

^{Â 2Â }= 12, Axis is x-axis

x

^{2}/4 â€“ y

^{2}/12 = 1

14. Find the equation of hyperbola satisfying the given condition:

Vertices (Â± 4, 0), e = 4/3

**Answer**

Veritices are (Â± 7, 0) âˆ´Â a = 7

e = c/a = 4/3 âˆ´Â c = 4/3 a = 4/3Â Ã—Â 7 = 28/3

b

^{2}Â = c

^{2}Â â€“ a

^{2}Â = (28/3)

^{2}Â â€“ 49 = 343/9

Axis is along x-axis

x

^{2}/49 - y

^{2}/343/9 = 1 or 7x

^{2}Â â€“ 9y

^{2}Â = 343

15. Find the equation of the hyperbola satisfying the given condition:

Foci (0, Â± âˆš10), passing through (2, 3)

**Answer**

Let the equation of hyperbola be

y

^{2}/b

^{2}- x

^{2}/a

^{2}Â = 1 â€¦(i)

âˆ´Â be =Â âˆš10

Also, a

^{Â 2}Â = b

^{Â 2}Â (e

^{Â 2}Â â€“ 1) = b

^{Â 2}Â e

^{Â 2}Â â€“ b

^{Â 2}Â = 10 â€“ b

^{Â 2}Â â€¦(ii)

Thus, the equation of the hyperbola is

y

^{2}/b

^{2}- x

^{2}/(10 â€“ b

^{2}) = 1

As, it passes through the point (2, 3)

âˆ´Â 9/b

^{2}- 4/(10 â€“ b

^{2}) = 1 => b

^{Â 4}Â â€“ 23b

^{Â 2}Â + 90 = 0

=> (b

^{Â 2}Â â€“ 5)(b

^{Â 2}Â â€“ 18) = 0 => b

^{Â 2}Â = 18, 5

When b

^{Â 2}Â = 18, then from (ii) a

^{Â 2}Â = â€“8 which is not possible and when b

^{Â 2}Â = 5, then from (ii) a

^{Â 2}Â = 5

Hence, the required equation of the hyperbola is y

^{2}/5 - x

^{2}/5 = 1 => y

^{2}Â - x

^{2}Â = 5