NCERT solutions of class 11 maths exercise 11.3 chapter 11-Conic Section

Here all NCERT Solutions of Class 11 Maths Exercise 11.3 Chapter 11 Are Solved For Helping The Students of Class 11 In Clearing The Concept Of Chapter 11 Conic Section Completely, So That They Could Attemp All The Questions Of The Chapter 11 Conic Section Mentioned In The Maths Question Paper Of CBSE Board Class 11.

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NCERT solutions of class 11 maths exercise 11.2 chapter 11-Conic sections

NCERT Solutions of class 11 maths exercise 11.1 chapter 11-Conic sections

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

$\boldsymbol{Q1.\frac{x^{2}}{36}+\frac{y^{2}}{16}=1}$

Ans. The given equation of the ellipse is

${\frac{x^{2}}{36}+\frac{y^{2}}{16}=1}$

The equation of ellipse shows that major axis of the ellipse is along x-axis

The standard equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

On Comparing the given equation with the standard equation,we get

a² = 36 ⇒ a = ±6 and b² = 16⇒b =±4

The coordinates of the foci are (±c,0)

c = √(a²-b²) = √(6² -4²) =√(36 -16) =√20

Therefore the coordinates of the foci are (±√20,0)

The vertices of the ellipse are (±a,0) = (±6,0)

The length of major axis is = 2a = 2× 6 = 12

The length of minor axis = 2b = 2×4 = 8

The eccentricity of ellipse is = e =c/a = √20/6

Length of the latus rectum of the ellipse = 2b²/a =2×4²/6=16/3

$\boldsymbol{Q2.{\frac{x^{2}}{4}+\frac{y^{2}}{25}=1}}$

Ans. The given equation of the ellipse is

${{\frac{x^{2}}{4}+\frac{y^{2}}{25}=1}}$

The equation of ellipse shows that major axis of the ellipse is along y-axis(25>4)

The standard equation of ellipse is

${{\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1}}$

On Comparing the given equation with the standard equation,we get

b² = 4 ⇒ b = ±2 and a² = 25⇒a =±5

The coordinates of the foci are (±c,0)

c = √(a²-b²) = √(5² -2²) =√(25 -4) =√21

Therefore the coordinates of the foci are (0,±√21)

The vertices of the ellipse are (0,±a) = (0,±5)

The length of major axis is = 2a = 2× 5 = 10

The length of minor axis = 2b = 2×2 = 4

The eccentricity of ellipse is = e =c/a = √21/5

Length of the latus rectum of the ellipse = 2b²/a =2×2²/5=8/5

$\boldsymbol{Q3.{{\frac{x^{2}}{16}+\frac{y^{2}}{9}=1}}}$

Ans. The given equation of the ellipse is

${{{\frac{x^{2}}{16}+\frac{y^{2}}{9}=1}}}$

The equation of ellipse shows that major axis of the ellipse is along x-axis(16>9)

The standard equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

On Comparing the given equation with the standard equation,we get

a² = 16 ⇒ a = ±4 and b² = 9⇒b=±3

The coordinates of the foci are (±c,0)

c = √(a²-b²) = √(4² -3²) =√(16 -9) =√7

Therefore the coordinates of the foci are (±√7,0)

The vertices of the ellipse are (±a,0) = (±4,0)

The length of major axis is = 2a = 2× 4 = 8

The length of minor axis = 2b = 2×3 = 6

The eccentricity of ellipse is = e =c/a = √7/4

Length of the latus rectum of the ellipse = 2b²/a =2×3²/4=9/2

$\boldsymbol{Q4.\frac{x^{2}}{25}+\frac{y^{2}}{100}=1}$

Ans. The given equation of the ellipse is

${\frac{x^{2}}{25}+\frac{y^{2}}{100}=1}$

The equation of ellipse shows that major axis of the ellipse is along y-axis(100>25)

The standard equation of ellipse is

${\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1}$

On Comparing the given equation with the standard equation,we get

a² = 100⇒ a = ±10 and b² = 25⇒b=±5

The coordinates of the foci are (0,±c)

c = √(a²-b²) = √(10² -5²) =√(100 -25) =√75= 5√3

Therefore the coordinates of the foci are (0,±5√3)

The vertices of the ellipse are (0,±a) = (0,±10)

The length of major axis is = 2a = 2× 10= 20

The length of minor axis = 2b = 2×5 = 10

The eccentricity of ellipse is = e =c/a = 5√3/10 =√3/2

Length of the latus rectum of the ellipse = 2b²/a =2×(5)²/10=5

$\boldsymbol{Q5.{\frac{x^{2}}{49}+\frac{y^{2}}{36}=1}}$

Ans. The given equation of the ellipse is

${{\frac{x^{2}}{49}+\frac{y^{2}}{36}=1}}$

The equation of ellipse shows that major axis of the ellipse is along y-axis(49>36)

The standard equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

On Comparing the given equation with the standard equation,we get

a² = 49⇒ a = ±7 and b² = 36⇒b=±6

The coordinates of the foci are (±c,0)

c = √(a²-b²) = √(7² -6²) =√(49-36) =√13

Therefore the coordinates of the foci are (±√13,0)

The vertices of the ellipse are (±a,0) = (±7,0)

The length of major axis is = 2a = 2× 7= 14

The length of minor axis = 2b = 2×6= 12

The eccentricity of ellipse is = e =c/a = √13/7 =√13/7

Length of the latus rectum of the ellipse = 2b²/a =2×(6)²/7=72/7

$\boldsymbol{{Q6.{\frac{x^{2}}{100}+\frac{y^{2}}{400}=1}}}$

Ans. The given equation of the ellipse is

${{{\frac{x^{2}}{100}+\frac{y^{2}}{400}=1}}}$

The equation of ellipse shows that major axis of the ellipse is along y-axis (400>100)

The standard equation of ellipse is

${{{\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1}}}$

On Comparing the given equation with the standard equation,we get

a² = 400⇒ a = ±20 and b² = 100⇒b=±10

The coordinates of the foci are (0,±c)

c = √(a²-b²) = √(400 -100) =√300 = 10√3

Therefore the coordinates of the foci are (0,±10√3)

The vertices of the ellipse are (0,±a) = (0,±20)

The length of major axis is = 2a = 2×20 = 40

The length of minor axis = 2b = 2×10= 20

The eccentricity of ellipse is = e =c/a = 10√3/20 =√3/2

Length of the latus rectum of the ellipse = 2b²/a =2×(10)²/20=10

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$\boldsymbol{Q7.36x^{2}+4y^{2}=144}$

Ans. The given equation of the ellipse is

${36x^{2}+4y^{2}=144}$

Rearranging the equation as follows

$\frac{36x^{2}}{144}+\frac{4y^{2}}{144}=1\Rightarrow \frac{x^{2}}{4}+\frac{y^{2}}{36}=1$

The equation of ellipse shows that major axis of the ellipse is along y-axis (36>4)

The standard equation of ellipse is

${{{\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1}}}$

On Comparing the given equation with the standard equation,we get

a² = 36⇒ a = ±6 and b² = 4⇒b=±2

The coordinates of the foci are (0,±c)

c = √(a²-b²) = √(6² -2²) =√(36-4) =√32= 4√2

Therefore the coordinates of the foci are (0,±4√2)

The vertices of the ellipse are (0,±a) = (0,±6)

The length of major axis is = 2a = 2× 6= 12

The length of minor axis = 2b = 2×2= 4

The eccentricity of ellipse is = e =c/a = 4√2/6 =2√2/3

Length of the latus rectum of the ellipse = 2b²/a =2×(2)²/6=4/3

$\boldsymbol{Q8.16x^{2}+y^{2}=16}$

Ans. The given equation of the ellipse is

Rearranging the equation as follows

$\frac{x^{2}}{1}+\frac{y^{2}}{16}=1$

The equation of ellipse shows that major axis of the ellipse is along y-axis (16>1)

The standard equation of ellipse is

${{{\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1}}}$

On Comparing the given equation with the standard equation,we get

a² = 16⇒ a = ±4 and b² = 1⇒b=±1

The coordinates of the foci are (0,±c)

c = √(a²-b²) = √(4² -1²) =√(16-1) =√15

Therefore the coordinates of the foci are (0,±√15)

The vertices of the ellipse are (0,±a) = (0,±4)

The length of major axis is = 2a = 2× 4= 8

The length of minor axis = 2b = 2×1= 2

The eccentricity of ellipse is = e =c/a = √15/4

Length of the latus rectum of the ellipse = 2b²/a =2×(1)²/4=1/2

$\boldsymbol{Q9.4x^{2}+9y^{2}=36}$

Ans. The given equation of the ellipse is

${4x^{2}+9y^{2}=36}$

Rearranging the equation as follows

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

The equation of ellipse shows that major axis of the ellipse is along y-axis (9>4)

The standard equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

On Comparing the given equation with the standard equation,we get

a² = 9⇒ a = ±3 and b² = 4⇒b=±2

The coordinates of the foci are (±c,0)

c = √(a²-b²) = √(3² -2²) =√(9-4) =√5

Therefore the coordinates of the foci are (±√5,0)

The vertices of the ellipse are (±a,0) = (±3,0)

The length of major axis is = 2a = 2× 3= 6

The length of minor axis = 2b = 2×2= 4

The eccentricity of ellipse is = e =c/a = √5/3

Length of the latus rectum of the ellipse = 2b²/a =2×(2)²/3=8/3

In each of the exercises 10 to 20,find the equations for the ellipse that satisfies the given conditions.

$\boldsymbol{Q10.Vertices (\pm 5,0),foci\: (\pm 4,0)}$

Ans. We are given the vertices (±a,0) = (±5,0), foci (±c,0) = (±4,0)

So,a = 5, c = 4

Since c² = a² – b² ⇒b =±√(a² -c²) =±√(5² – 4²) = ±√(25 -16) =±√9=±3

The equation of equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting the value of a = 5 and b =3 in the equation

$\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$

$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Q11. Vertices (0, ±13), foci (0, ± 5)

Ans. We are given the vertices (0,±a) = (0,±13), foci (0, ±c) = (0,±5)

So,a = 13, c = 5

Since c² = a² – b² ⇒b =±√(a² -c²) =±√(13² – 5²) = ±√(169 -25) =±√144=±12

The equation of equation of ellipse is

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$

Putting the value of a = 13 and b =12 in the equation

$\frac{x^{2}}{12^{2}}+\frac{y^{2}}{13^{2}}=1$

$\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$

Q12. Vertices (±6,0), foci ( ± 4,0)

Ans. We are given the vertices (±a,0) = (±6,0), foci ( ±c,0) = (±4,0)

So,a = 6, c = 4

Since c² = a² – b² ⇒b =±√(a² -c²) =±√(6² – 4²) = ±√(36 -16) =±√20=±2√5

The equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting the value of a = 6 and b =2√5 in the equation

$\frac{x^{2}}{6^{2}}+\frac{y^{2}}{\left ( 2\sqrt{5} \right )^{2}}=1$

$\frac{x^{2}}{36}+\frac{y^{2}}{20}=1$

Q13. Ends of the major axis (±3,0), ends of the minor axis (0, ± 2)

Ans. We are given the vertices i.e ends of the major axis (±a,0) = (±3,0), ends of the minor axis i.e (0,b) = (0, ± 2)

So,a = 3, b= 2

The equation of equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

$\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=1$

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

Q14. Ends of the major axis (0,±√5), ends of the minor axis (± 1,0)

Ans. We are given the vertices i.e ends of the major axis (0, ±a) = (0,±√5), ends of the minor axis i.e (±b, 0) = (± 1,0)

So,a = √5, b= 1

The equation of the ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

$\frac{x^{2}}{\sqrt{5}^{2}}+\frac{y^{2}}{1^{2}}=1$

$\frac{x^{2}}{5}+\frac{y^{2}}{1}=1$

Q15. Length of major axis 26, foci (±5, 0)

Ans.We are given length of major axis 26, foci (±5, 0)

Length of major axis is 26 and foci (±5, 0)

Length of major axis, 2a = 26 ⇒ a = 13

Foci, (±c,0) = (±5, 0)⇒ c = 5

It is known that a² = b²+ c².⇒b =±√(a²-c²) = ±√(13² – 5²)= ±√(169 – 25)=±√144=±12⇒b =12

The equation of the ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting the value of a =13, b = 12

$\frac{x^{2}}{13^{2}}+\frac{y^{2}}{12^{2}}=1$

$\frac{x^{2}}{169}+\frac{y^{2}}{144}=1$

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

Ans.We are given length of minor axis 16, foci (0,±6)

Length of minor axis, 2b = 16 ⇒ b = 8

Foci, (0, ±c) = (0, ±6)⇒ c = 6

It is known that a² = b²+ c².⇒a =±√(b²+c²) = ±√(8² + 6²)= ±√(64 + 36)=±√100=±10⇒a =10

The equation of the ellipse is

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$

Putting the value a =10,b =8

∴ The equation of the ellipse

$\frac{x^{2}}{8^{2}}+\frac{y^{2}}{10^{2}}=1$

$\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$

Q17.Foci (±3, 0), a = 4

Ans. We are given that

Foci (±3, 0) and a = 4

Foci, ( ±c,0) = ( ±3,0)⇒ c = 3

It is known that a² = b²+ c².⇒b =±√(a²-c²) = ±√(4² – 3²)= ±√(16- 9)=±√7=±√7⇒a =√7

The equation of ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting b = √7 and a = 4

$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$

Q18. b = 3, c = 4, centre at the origin; foci on the x axis.

Ans. We are given that

b = 3, c = 4, centre is(0,0) and foci on the x axis =(±c,0) = (±3,0)

It is known that a² = b² + c² ⇒ a =±√(b² +c²) = ±√(3² +4²) =±√25=±5

The equation of the ellipse is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting the value of a =5 and b =3

$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Q19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Ans. We are given centre of ellipse at (0,0) and it is passing through (3, 2) and (1, 6)

Major axis is on y-axis,so the eqation is

$\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$

Putting the value of (x,y)= (3, 2) and (x,y) =(1, 6) in the equation,we get the equation (i) and equation (ii)

$\frac{9}{b^{2}}+\frac{4}{a^{2}}=1.....\left ( i \right )$

$\frac{1}{b^{2}}+\frac{36}{a^{2}}=1.....\left ( ii \right )$

Solving (i) and (ii) equations

b² = 10 and a²= 40.

Substituting the value of b² = 10 and a²= 40 in the equation,we get the required equation

$\frac{x^{2}}{10}+\frac{y^{2}}{40}=1$

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

Ans. We are given that the ellipse is passing through (4, 3) and ( 6,2)

Major axis is on x-axis,so the eqation is

${\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1}$

Putting the value of (x,y)= (4, 3) and (x,y) =(6, 2) in the equation,we get the equation (i) and equation (ii)

$\frac{16}{a^{2}}+\frac{9}{b^{2}}=1.....\left ( i \right )$

$\frac{36}{a^{2}}+\frac{4}{b^{2}}=1.....\left ( ii \right )$

Solving (i) and (ii) equations

b² = 13 and a²= 52.

Substituting the value of b² = 13 and a²= 52 in the equation,we get the required equation

$\frac{x^{2}}{52}+\frac{y^{2}}{13}=1$

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

Solutions of Class 11 maths test unit-2 G.D Lancer Public School,New Delhi

NCERT Solutions of Class 11 Maths Exercise 9.3 of the chapter 9-Sequences and Series

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NCERT solutions of class 11 maths exercise 11.3 chapter 11-Conic Section

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