# Class 11 NCERT Solutions- Chapter 11 Conic Section – Exercise 11.3

### In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse.

### Question 1. = 1

**Solution:**

Since denominator of x

^{2}/36 is larger than the denominator of y^{2}/16,the major axis is along the x-axis.

On comparing the given equation with = 1, we get

a

^{2 }= 36 and b^{2 }= 16⇒ a = ±6 and b = ±4

The Foci:Foci = (c, 0) and (-c, 0) when (a

^{2 }> b^{2})c = √(a

^{2 }– b^{2}) -(when a^{2 }> b^{2})c = √(36 – 16)

c = √20 = 2√5

⇒ (2√5, 0) and (-2√5, 0)

Vertices:Vertices = (a, 0) and (-a, 0) when (a

^{2 }> b^{2})⇒ (6, 0) and (-6, 0)

Length of major axis:Length of major axis = 2a (when a

^{2 }> b^{2})= 2 × 6

⇒ Length of major axis = 12

Length of minor axis:Length of minor axis = 2b (when a

^{2 }> b^{2})= 2 × 4

⇒ Length of minor axis = 8

EccentricityEccentricity = c/a (when a

^{2 }> b^{2})= 2√5/6

= √5/3

Length of the latus rectum:Length of the latus rectum = 2b

^{2}/a (when a^{2 }> b^{2})= 2×16/6

= 16/3

### Question 2. = 1

**Solution:**

Since denominator of y

^{2}/25 is larger than the denominator of x^{2}/4,the major axis is along the y-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 4 and b^{2 }= 25⇒ a = ±2 and b = ±5

The Foci:Foci = (0, c) and (0, -c) when (a

^{2 }< b^{2})c = √(b

^{2 }– a^{2}) -(when a^{2 }< b^{2})c = √(25 – 4)

c = √21

⇒ (0, √21) and (0, -√21)

Vertices:Vertices = (0, b) and (0, -b) when (a

^{2 }< b^{2})⇒ (0, 5) and (0, -5)

Length of major axis:Length of major axis = 2b (when a

^{2 }< b^{2})= 2 × 5

⇒ Length of major axis = 10

Length of minor axis:Length of minor axis = 2a (when a

^{2 }< b^{2})=2 × 2

⇒ Length of minor axis =

4

Eccentricity:Eccentricity = c/b (when a

^{2 }< b^{2})= √21/5

Length of the latus rectum:Length of the latus rectum = 2a

^{2}/b (when a^{2 }< b^{2})= 2×4/5

= 8/5

### Question 3. = 1

**Solution:**

Since denominator of x

^{2}/16 is larger than the denominator of y^{2}/9,the major axis is along the x-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 16 and b^{2 }= 9⇒ a = ±4 and b = ±3

The Foci:Foci = (c, 0) and (-c, 0) when (a

^{2 }> b^{2})c = √(a

^{2 }– b^{2}) -(when a^{2 }> b^{2})c = √(16 – 9)

c = √7

⇒ (√7, 0) and (-√7, 0).

Vertices:Vertices = (a, 0) and (-a, 0) when (a

^{2 }> b^{2})⇒ (4,0) and (-4,0)

Length of major axis:Length of major axis = 2a (when a

^{2 }> b^{2})= 2 × 4

⇒ Length of major axis = 8

Length of minor axis:Length of minor axis = 2b (when a

^{2 }> b^{2})= 2 × 3

⇒ Length of minor axis = 6

Eccentricity:Eccentricity = c/a (when a

^{2}>b^{2})= √7/4

Length of the latus rectum:Length of the latus rectum = 2b

^{2}/a (when a^{2 }> b^{2})= 2 × 9/4

= 9/2

### Question 4. = 1

**Solution:**

Since denominator of y

^{2}/100 is larger than the denominator of x^{2}/25,the major axis is along the y-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 25 and b^{2 }= 100⇒ a = ±5 and b = ±10

The Foci:Foci = (0, c) and (0, -c) when (a

^{2 }< b^{2})c = √(b

^{2 }– a^{2}) -(when a^{2 }< b^{2})c = √(100 – 25)

c = √75

c = 5√3

⇒ (0, 5√3) and (0, -5√3)

Vertices:Vertices = (0, b) and (0, -b) when (a

^{2 }< b^{2})⇒ (0, 10) and (0, -10)

Length of major axis:Length of major axis = 2b (when a

^{2 }< b^{2})= 2 × 10

⇒ Length of major axis = 20

Length of minor axis:Length of minor axis = 2a (when a

^{2 }< b^{2})= 2 × 5

⇒ Length of minor axis = 10

Eccentricity:Eccentricity = c/b (when a

^{2 }< b^{2})= 5√3/10

= √3/2

Length of the latus rectum:Length of the latus rectum = 2a

^{2}/b (when a^{2 }< b^{2})= 2 × 25/10

= 5

### Question 5. = 1

**Solution:**

Since denominator of x

^{2}/49 is larger than the denominator of y^{2}/36,the major axis is along the x-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 49 and b^{2 }= 36⇒ a = ±7 and b = ±6

The Foci:Foci = (c, 0) and (-c, 0) when (a

^{2 }> b^{2})c = √(a

^{2 }– b^{2}) -(when a^{2 }> b^{2})c = √(49 – 36)

c = √13

⇒ (√13, 0) and (-√13, 0).

Vertices:Vertices = (a, 0) and (-a, 0) when (a

^{2 }> b^{2})⇒ (7, 0) and (-7, 0)

Length of major axis:Length of major axis = 2a (when a

^{2 }> b^{2})= 2 × 7

⇒ Length of major axis = 14

Length of minor axis:Length of minor axis = 2b (when a

^{2 }> b^{2})= 2 × 6

⇒ Length of minor axis = 12

Eccentricity:Eccentricity = c/a (when a

^{2 }> b^{2})= √13/7

Length of the latus rectum:Length of the latus rectum = 2b

^{2}/a (when a^{2 }> b^{2})= 2 × 36/7

= 72/7

### Question 6. = 1

**Solution:**

Since denominator of y

^{2}/400 is larger than the denominator of x^{2}/100,the major axis is along the y-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 100 and b^{2 }= 400⇒ a = ±10 and b = ±20

The Foci:Foci = (0, c) and (0, -c) when (a

^{2 }< b^{2})c = √(b

^{2 }– a^{2}) -(when a^{2 }< b^{2})c = √(400 – 100)

c = √300

c = 10√3

⇒ (0, 10√3) and (0, -10√3)

Vertices:Vertices = (0, b) and (0, -b) when (a

^{2 }< b^{2})⇒ (0, 20) and (0, -20)

Length of major axis:Length of major axis = 2b (when a

^{2 }< b^{2})= 2 × 20

⇒ Length of major axis = 40

Length of minor axis:Length of minor axis = 2a (when a

^{2 }< b^{2})= 2 × 10

⇒ Length of minor axis = 20

Eccentricity:Eccentricity = c/b (when a

^{2 }< b^{2})= 10√3/20

= √3/2

Length of the latus rectum:Length of the latus rectum = 2a

^{2}/b (when a^{2 }< b^{2})= 2×100/20

= 10

### Question 7. 36x^{2} + 4y^{2} = 144

**Solution:**

36x

^{2}+ 4y^{2}= 144Dividing LHS and RHS by144,

= 1 (Obtained Equation)

Since denominator of y

^{2}/36 is larger than the denominator of x^{2}/4,the major axis is along the y-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 4 and b^{2 }= 36⇒ a = ±2 and b = ±6

The Foci:Foci = (0, c) and (0, -c) when (a

^{2 }< b^{2})c = √(b

^{2 }– a^{2}) -(when a^{2 }< b^{2})c = √(36 – 4)

c = √32

c = 4√2

⇒ (0, 4√2) and (0, -4√2)

Vertices:Vertices = (0, b) and (0, -b) when (a

^{2 }< b^{2})⇒ (0, 6) and (0, -6)

Length of major axis:Length of major axis = 2b (when a

^{2}<b^{2})= 2 × 6

⇒ Length of major axis = 12

Length of minor axisLength of minor axis = 2a (when a

^{2 }< b^{2})= 2 × 2

⇒ Length of minor axis = 4

Eccentricity:Eccentricity = c/b (when a

^{2 }< b^{2})= 4√2/6

= 2√2/3

Length of the latus rectum:Length of the latus rectum = 2a

^{2}/b (when a^{2 }< b^{2})= 2 × 4/6

= 4/3

### Question 8. 16x^{2} + y^{2} = 16

**Solution:**

16x

^{2}+ y^{2}= 16Dividing LHS and RHS by16,

= 1 (Obtained Equation)

Since denominator of y

^{2}/16 is larger than the denominator of x^{2}/1,the major axis is along the y-axis.

Comparing the given equation with = 1, we get

a

^{2 }= 1 and b^{2 }= 16⇒ a = ±1 and b = ±4

The Foci:Foci = (0, c) and (0, -c) when (a

^{2 }< b^{2})c = √(b

^{2 }– a^{2}) -(when a^{2 }< b^{2})c = √(16 – 1)

c = √15

⇒ (0, √15) and (0, -√15)

Vertices:Vertices = (0, b) and (0, -b) when (a

^{2 }< b^{2})⇒ (0, 4) and (0, -4)

Length of major axis:Length of major axis = 2b (when a

^{2 }< b^{2})= 2 × 4

⇒ Length of major axis = 8

Length of minor axis:Length of minor axis = 2a (when a

^{2 }< b^{2})=2 × 1

⇒ Length of minor axis = 2

Eccentricity:Eccentricity = c/b (when a

^{2 }< b^{2})= √15/4

Length of the latus rectum:Length of the latus rectum = 2a

^{2}/b (when a^{2 }< b^{2})= 2 × 1/4

= 1/2

### Question 9. 4x^{2} + 9y^{2} = 36

**Solution:**

4x

^{2}+ 9y^{2}= 36Dividing LHS and RHS by 36,

= 1 (Obtained Equation)

Since denominator of x

^{2}/9 is larger than the denominator of y^{2}/4,the major axis is along the x-axis.

Comparing the given equation with = 1, we get

a

^{2}= 9 and b^{2}= 4⇒ a = ±3 and b = ±2

The Foci:Foci = (c, 0) and (-c, 0) when (a

^{2}> b^{2})c = √(a

^{2}– b^{2}) -(when a^{2}> b^{2})c = √(9 – 4)

c = √5

⇒ (√5, 0) and (-√5, 0).

VerticesVertices = (a, 0) and (-a, 0) when (a

^{2}> b^{2})⇒ (3, 0) and (-3, 0).

Length of major axisLength of major axis = 2a (when a

^{2}> b^{2})= 2 × 3

⇒ Length of major axis = 6

Length of minor axisLength of minor axis = 2b (when a

^{2}> b^{2})= 2 × 2

⇒ Length of minor axis = 4

EccentricityEccentricity = c/a (when a

^{2}> b^{2})= √5/3

Length of the latus rectumLength of the latus rectum = 2b

^{2}/a (when a^{2}> b^{2})= 2 × 4/3

= 8/3

### In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

### Question 10. Vertices (± 5, 0), foci (± 4, 0).

**Solution:**

Since the vertices are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that a = ±5, c = ±4

As, from the relation

c

^{2}= a^{2 }– b^{2}(when a^{2 }> b^{2})b

^{2}= a^{2 }– c^{2}b

^{2}= 25 – 16b

^{2}= 9So, a

^{2}= 25 and b^{2}= 9Hence, the required equation of ellipse,

= 1

### Question 11. Vertices (0, ± 13), foci (0, ± 5).

**Solution:**

Since the vertices are on y-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a

^{2 }< b^{2})Given that b = ±13, c = ±5

As, from the relation

c

^{2}= b^{2 }– a^{2}(when a^{2 }< b^{2})a

^{2}= b^{2 }– c^{2}a

^{2}= 169 – 25a

^{2}= 144So, a

^{2}= 144 and b^{2}= 169Hence, the required equation of ellipse,

= 1

### Question 12. Vertices (± 6, 0), foci (± 4, 0).

**Solution:**

Since the vertices are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that a = ±6, c = ±4

As, from the relation

c

^{2}= a^{2 }– b^{2}(when a^{2 }> b^{2})b

^{2}= a^{2 }– c^{2}b

^{2}= 36 – 16b

^{2}= 20So,

a^{2}= 36 and b^{2}= 20Hence, the required equation of ellipse,

= 1

### Question 13. Ends of major axis (± 3, 0), ends of minor axis (0, ± 2).

**Solution:**

Since the major axis are on x-axis, and minor axis on the y-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that a = ±3, b = ±2

So, a

^{2}= 9 and b^{2}= 4Hence, the required equation of ellipse,

= 1

### Question 14. Ends of major axis (0, ±√5), ends of minor axis (± 1, 0).

**Solution:**

Since the major axis are on y-axis, and minor axis on the x-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a

^{2 }< b^{2})Given that a = ±1, b = ±√5

So, a

^{2}= 1 and b^{2}= 5Hence, the required equation of ellipse,

= 1

### Question 15. Length of major axis 26, foci (± 5, 0).

**Solution:**

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that c = ±5 and Length of major axis = 26

As, Length of major axis = 2a (when a

^{2}> b^{2})2a = 26

a = 13

As, from the relation

c

^{2}= a^{2 }– b^{2}(when a^{2 }> b^{2})b

^{2}= a^{2 }– c^{2}b

^{2}= 169 – 25b

^{2}= 144So, a

^{2}= 169 and b^{2}= 144Hence, the required equation of ellipse,

= 1

### Question 16. Length of minor axis 16, foci (0, ± 6).

**Solution:**

Since the foci are on y-axis, the equation will be of the form

= 1, where b is the semi-major axis. (where a

^{2 }< b^{2})Given that c = ±6 and Length of minor axis = 16

As, Length of minor axis = 2a (when a

^{2 }< b^{2})2a = 16

a = 8

As, from the relation

c

^{2}= b^{2 }– a^{2}(when a^{2 }< b^{2})b

^{2}= c^{2 }+ a^{2}b

^{2}= 36 + 64b

^{2}= 100So, a

^{2}= 64 and b^{2}= 100Hence, the required equation of ellipse,

= 1

### Question 17. Foci (± 3, 0), a = 4.

**Solution:**

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that a = 4 and c = ±3

As, from the relation

c

^{2}= a^{2 }– b^{2}(when a^{2 }> b^{2})b

^{2}= a^{2 }– c^{2}b

^{2}= 16 – 9b

^{2}= 7So, a

^{2}= 16 and b^{2}= 7Hence, the required equation of ellipse,

= 1

### Question 18. b = 3, c = 4, centre at the origin; foci on the x axis.

**Solution:**

Since the foci are on x-axis, the equation will be of the form

= 1, where a is the semi-major axis. (where a

^{2 }> b^{2})Given that b = 3 and c = 4

As, from the relation

c

^{2}= a^{2 }– b^{2}(when a^{2 }> b^{2})a

^{2}= b^{2 }+ c^{2}a

^{2}= 9 + 16a

^{2}= 25So, a

^{2}= 25 and b^{2}= 9Hence, the required equation of ellipse,

= 1

### Question 19. Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

**Solution:**

The standard equation of ellipse having centre (0, 0) will be of the form

= 1

Since the points (3, 2) and (1, 6) lie on the ellipse, we can have

= 1

= 1 -(1)

= 1

= 1 -(2)

Eq(2) subtracted from (multiplying eq(1) by 9) and, we get

9×() – () = 9 – 1

^{ }= 880/a

^{2}= 8a

^{2}= 80/8a

^{2}= 10Now, substituting a

^{2}= 10 in eq(1)= 1

9/10 + 4/b

^{2}= 14/b

^{2}= 1 – 9/104/b

^{2}= 1/10b

^{2}= 10 × 4 = 40So, a

^{2}= 10 and b^{2}= 40Hence, the required equation of ellipse,

= 1

### Question 20. Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

**Solution:**

The standard equation of ellipse having centre (0, 0) will be of the form

= 1

Since the points (4,3) and (6,2) lie on the ellipse, we can have

= 1

= 1 -(1)

and, = 1

= 1 -(2)

(multiplying eq(2) by 9) subtracted from (multiplying eq(1) by 4) and, we get

9×( ) – 4×() = 9 – 4

= 5

260/a

^{2}= 5a

^{2}= 260/5a

^{2}= 52Now, substituting a

^{2}= 52 in eq(1)= 1

9/b

^{2}= 1 – 16/529/b

^{2}= 36/529/b

^{2}= 36/52b

^{2}= 9 × 36/52 = 13So, a

^{2}= 52 and b^{2}= 13Hence, the required equation of ellipse,

= 1