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Edited By Ravindra Pindel | Updated on Sep 12, 2022 04:13 PM IST

**NCERT Exemplar Class 11 Maths Solutions Chapter 11** have been formulated by the professionals for the students to get detailed yet interesting insight into the concept of coordinate geometry in reference to curves. The chapter explains and covers a wider section of coordinate geometry dealing with the concept of curves through the previously learned concepts of geometry and algebra dealing with conic sections of circle, ellipse, parabola and hyperbola. NCERT exemplar Class 11 Maths solutions chapter 11 are provided by the professionals to provide efficient study material and help to students in order to assist them perfectly for any form of examination and application of the concept in different situations. The experts have used a very easy yet powerful approach for the NCERT exemplar Class 11 Maths solutions chapter 11 that make the learning process interesting yet knowledgeable.**Also, check -** NCERT Class 11 Maths Solutions for Other Chapters

**JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | **

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**Scholarship Test:** **Vidyamandir Intellect Quest (VIQ)**

Question:1

Find the equation of the circle which touches both axes in first quadrant and whose radius is a.

Answer:

Given that the circle has a radius aand touches both axis. So the centre is (a,a).

(x-a)^{2}+(y-a)^{2}=a^{2}

x^{2}+y^{2}-2ax-2ay+a^{2}=0

Question:2

Answer:

We have variable point as and

Question:3

If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.

Answer:

We have a circle through the point A(0,0), B(a,0) and C(0,b). Clearly triangle is right angled at vertex A. So, the centre of the circle is the mid-point of hypotenuse BC which is (a/2,b/2)

Question:4

Find the equation of the circle which touches x-axis and whose centre is (1, 2).

Answer:

It is given that circle with centre (1,2) touches x axis

Radius of the circle is r = 2

So equation of the required circle is

(x-1)^{2}+(y-2)^{2}=(2)^{2}

x^{2}-2x+1+y^{2}-4y+4=4

x^{2}+y^{2}-2x-4y+1=0

Question:5

Answer:

Given lines are 6x-8y+8=0 and 6x-8y-7=0

These parallel lines are tangent to a circle

Diameter of circle = Distance between the lines

Diameter of circle=

Radius of circle=

Question:6

Answer:

Since circle touches both the axes, its centre is C(-a, -a) and radius is a

Also, circle touches the line 3x-4y+8=0

Distance from centre C to this line is radius of the circle

Radius of circle,

a=2 or a=-4/3

a=2

Equation of required circle

(x+2)^{2}+(y+2)^{2}=2^{2}

x^{2}+y^{2}+4x+4y+4=0

Question:7

Answer:

Given equation of the circle is

x^{2}+y^{2}-4x-6y+11=0

2g=-4 and 2f=-6

So centre of circle is C(-g,-f) = C(2,3)

A (3,4) is one end of diameter . Let the other end of the diameter be B (x_{1}, y_{1})

and

x_{1}=1 and y_{1}=2

Question:8

Answer:

Given lines are

3x+y=14 and 2x+5y=18

Solving these equations we get point of intersection of the lines as A (4,2)

Radius=

So, equation of the required circle is:

(x-1)^{2}+(y+2)^{2}=5^{2}

x^{2}+y^{2}-2x+4y-20=0

Question:9

If the line touches the circle x^{2} + y^{2} = 16, then find the value of k.

Answer:

Given line is and the circle is

x^{2}+y^{2}=16

Centre of the circle is (0,0) and radius is 4.

Since the line touches the circle, perpendicular distance from (0,0) to line

is equal to the radius of the circle.

k=±8

Question:10

Answer:

Given equation of the circle is :

x^{2}+y^{2}-6x+12y+15=0

(x-3)^{2}+(y+6)^{2}=(30)

Hence, centre is (3, -6) and radius is

Since, the required circle is concentric with above circle, centre of the required circle is (3, -6).

Let its radius be r

Area of circle=

r^{2}=60

Equation of required circle (x-3)^{2}+(y+6)^{2}=60

x^{2}+y^{2}-6x+12y-15=0

Question:11

If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Answer:

Consider the equation of the ellipse as

It is given that length of latus rectum = Half of minor axis

a=2b

b^{2}=a^{2}(1-e^{2})

b^{2}=4b^{2}(1-e^{2})

1-e^{2}=1/4

e^{2}=3/4

Question:12

Given the ellipse with equation 9x^{2} + 25y^{2} = 225, find the eccentricity and foci.

Answer:

Given equation of ellipse

9x^{2}+25y^{2}=225

a=5, b=3

b^{2}=a^{2}(1-e^{2})

9=25(1-e^{2})

Question:13

Answer:

Let the equation of the ellipse be

Distance between foci 2ae=10

It si given that eccentricity

Length of latus rectum of ellipse=

Question:14

Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0, 0).

Answer:

Let the equation of the ellipse be

Given that latus rectum

The required equation of the ellipse is

Question:15

Find the distance between the directrices of the ellipse

Answer:

The equation of the ellipse is

a=6 and

We know that

Distance between directrix=

Question:16

Find the coordinates of a point on the parabola y^{2 }= 8x whose focal distance is 4.

Answer:

Given parabola is y^{2}=8x

On comparing this parabola to the y^{2}=4ax

we get a=2

Focal distance=Distance of any point on parabola from the focus SP=

|x_{1}+2|=4

x_{1}=2,-6

But x≠ -6

So for x=2

y= ±4 So the points are (2,4) and 2,-4

Question:17

Answer:

Let the point on the parabola be P (x_{1},y_{1})

Slope of OP=

Question:18

Answer:

Given that the vertex of the parabola is A(0,4) and its focus is S(0,2)

So, directrix of the parabola is y=6

Now by definition of the parabola for any point P (x,y) on the parabola SP=PM

x^{2}+y^{2}-4y+4=y^{2}-12y+36

x^{2}+8y=32

Question:19

If the line y = mx + 1 is tangent to the parabola y^{2} = 4x then find the value of m.

Answer:

Given that line y=mx+1 is tangent to the parabola y^{2}=4x

Solving line with parabola we have

(mx+1)^{2}=4x

m^{2}x^{2}+2mx+1=4x

m^{2}x^{2}+x(2m-4)+1=0

Since the line touches the parabola, above equation must have equal roots.

Discriminant D=0

(2m-4)^{2}-4m^{2}=0

4m^{2}-16m+16-4m^{2}=0

16m=16

m=1

Question:20

Let the equation of the hyperbola be

Distance between foci=2ae=16

We know that

x^{2}-y^{2}=32

Question:21

Find the eccentricity of the hyperbola 9y^{2} – 4x^{2} = 36

Answer:

We have the hyperbola 9y^{2}-4x^{2}=36

a^{2}=9

b^{2}=4

a^{2}=b^{2}(e^{2}-1)

Question:22

Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Answer:

Let the equation of the hyperbola be

ae=2

b^{2}=a^{2}(e^{2}-1)

Question:23

Answer:

Given that the lines 2x-3y-5=0 and 3x-4y-7=0 are diameters of the circle.

2x-3y-5=0.....(i)

3x-4y-7=0....(ii)

3 (Equation (i)) - 2 (Equation (ii) )

6x-9y-15-6x+8y+14=0.

y = -1

Put value of y in equation (i)

x = 1

Solving these lines we will get the intersection as (1,-1) which is centre of the circle.

Also, given that area of the circle is 154 sq units

πr^{2}=154

r^{2}=154*7/22=49

r=7

(x-1)^{2}+(y+1)^{2}=49

x^{2}+y^{2}-2x+2y-47=0

Question:24

Answer:

Let the centre of the circle be C(h,k)

Given that the centre lies on the line y-4x+3=0

y-4x+3=0

k-4h+3=0

k=4h-3

AC^{2}=BC^{2}

(h-2)^{2}+(4h-3-3)^{2}=(h-4)^{2}+(4h-3-5)^{2}

h^{2}-4h+4+16h^{2}-48h+36=h^{2}-8h+16+16h^{2}-64h+64

20h=40

h=2

k=4h-3=5

radius=

Therefore equation of the circle is : (x-2)^{2}+(y-5)^{2}=4

x^{2}+y^{2}-4x-10y+25=0

Question:25

Answer:

Given centre of the circle O(3, -1)

Chord of the circle is AB

Given that equation of AB is 2x-5y+18=0

Perpendicular distance from O to AB is OP=

OB^{2}=OP^{2}+PB^{2}

OB^{2}=29+9=38

OB=

Equation of the circle is (x-3)^{2}+(y+1)^{2}=38

x^{2}+y^{2}-6x+2y=28

Question:26

Answer:

Given circle is x^{2}+y^{2}-2x-4y=20

(x-1)^{2}+(y-2)^{2}=5^{2}

Centre of this circle is C_{1}(1,2).

Now, the required circle of radius'5'touches the above circle at P(5,5).

Let the centre of the required circle be C_{2}(h,k).

Since the radius of the given circle and the required circle is same, point P is mid point of C_{1}C_{2}.

and

h=9 and k=8

So the equation of the required circle is (x-9)^{2}+(y-8)^{2}=25

x^{2}+y^{2}-18x-16y+120=0

Question:27

Answer:

Given that circle passes through the point A(7,3) and its radius is 3

Therefore centre of the circle is C(h,h-1)

Now, radius of the circle is AC=3

(h-7)^{2}+(h-1-3)^{2}=3^{2}

2h^{2}-22h+56=0

h^{2}-11h+28=0

(h-4)(h-7)=0

h=4 or 7

If the centre of the circle is C(4,3)⇒(x-4)^{2}+(y-3)^{2}=9⇒x^{2}+y^{2}-8x-6y+16=0

If the centre is C(7,6)⇒(x-7)^{2}+(y-6)^{2}=9⇒x^{2}+y^{2}-14x-12y+76=0

Question:28

Answer:

We know that the distance of any point on the parabola from its focus and its directrix is same.

i) Given that directrix x=0 and focus (6,0)

So, for any point P(x,y) on the parabola

Distance of P from directrix=Distance of P from focus x^{2}=(x-6)^{2}+y^{2}

y^{2}-12x+36=0

ii) Given that vertex=(0,4) and focus (0,2)

Now distance between the vertex and directrix is same as the distance between the vertex and focus.

Distance of P from directrix=Distance of P from focus

y^{2}-12y+36=x^{2}+y^{2}-4y+4

x^{2}=32-8y

iii) Given that focus is at (-1,-2)

and directrix x-2y+3=0

x^{2}+2x+1+y^{2}+4y+4=1/5[x^{2}+4y^{2}+9+6x-4xy-12y]

4x^{2}+4xy+y^{2}+4x+32y+16=0

Question:29

Answer:

Let the coordinates of the variable point be (x,y)

Then according to the question

x^{2}-6x+9+y^{2}=

Again squaring both sides, we get

x^{2}-36x+324=4(x^{2}-18x+81+y^{2})

3x^{2}+4y^{2}-36x=0 which is an ellipse.

Question:30

Answer:

Let the point be P (x,y)

According to the question

9(x^{2}+y^{2}-8y+16)=4(y^{2}-18y+81)

9x^{2}+5y^{2}=180

Question:31

Answer:

Let the point be P (x,y)

According to the question

Distance of P from (-4,0) -Distance of P from (4,0)=2

Squaring both the sides

16x^{2}-8x+1=x^{2}+16-8x+y^{2}

15x^{2}-y^{2}=15

Which is an equation of a hyperbola.

Question:32

Answer:

a) Given that a=5 and ae=7

e=7/5

b^{2}=a^{2(}e^{2}-1)=25(49/25-1)=24

So, the equation of hyperbola is

b) b=7,e=4/3

c) Given that foci=(0,±10 )

a^{2}=b^{2}(e^{2}-1)

a^{2}=b^{2}e^{2}-b^{2}

a^{2}=10-b^{2}

Since, hyperbola passes through the point (2,3)

4b^{2}-9(10-b^{2})=-b^{2}(10-b^{2})

b^{4}-23b^{2}+90=0

b^{4}-18b^{2}-5b^{2}+90=0

(b^{2}-18)(b^{2}-5)=0

b^{2}=5

a^{2}+b^{2}=10⇒a^{2}=5

y^{2}-x^{2}=5

Question:33

The line x + 3y = 0 is a diameter of the circle x^{2} + y^{2} + 6x + 2y = 0.

Answer:

False

Given equation of the circle is x^{2}+y^{2}+6x+2y=0

Centre=(-3,-1)

Clearly, it does not lie on the line x+3y=0 as-3+3(-1)=-6

So this line is not the diameter of the circle

Question:35

Answer:

True

Given circle is x^{2}+y^{2}=a^{2}

The given line lx+my-1=0 is tangent to the circle

Distance of (0,0) from the line lx+my-1=0 is equal to the radius a

The locus of (l,m) is . which is an equation of a circle.

Question:36

The point (1, 2) lies inside the circle x^{2} + y^{2} – 2x + 6y + 1 = 0.

Answer:

False

Given circle is x^{2}+y^{2}-2x+6y+1=0 or (x-1)^{2}+(y+3)^{2}=3^{2}

Centre is C(1,-3) and radius is 3.

Distance of point P (1,2) from centre is 5.

Thus CP>radius

So, point P lies outside the circle.

Question:37

The line lx + my + n = 0 will touch the parabola y^{2} = 4ax if ln = am^{2}.

Answer:

True

Give line lx+my+n=0

and parabola y^{2}=4ax

Solving line and parabola for their point of intersection we get

Since line touches the parabola, above equation must have equal roots

D=0

am^{2}=nl

Question:38

If P is a point on the ellipse whose foci are S and S′, then PS + PS′ = 8.

Answer:

False

We have definition of the ellipse is

From the definition of the ellipse, we know that sum of the distances of any point P

on the ellipse from the two foci is equal to the length of the major axis.

Here major axis=2b=2*5=10

S and S'are foci, then SP+S'P=10

Question:39

The line 2x + 3y = 12 touches the ellipse at the point (3, 2).

Answer:

True

Given line is 2x+3y=12

and the ellipse 4x^{2}+9y^{2}=72

Solving line and ellipse we get (12-3y)^{2}+9y^{2}=72

(4-y)^{2}+y^{2}=8

2y^{2}-8y+8=0

y^{2}-4y+4=0

(y-2)^{2}=0

y=2;x=3

Question:40

Answer:

True

Given equation of lines are

and

3x^{2}-y^{2}=48

which is the equation of hyperbola

a^{2}=16 and b^{2}=48

e^{2}=1+48/16=4

e=2

Question:41

Answer:

The perpendicular distance from centre (3,-4) to the given line is,

= which is radius of the circle

Required equation of the circle is

Question:42

Answer:

Given equation of line are y=x+2

3y=4x

2y=3x

Solving these lines, we get points of intersection A(6,8), B (4,6) and C(0,0).

Let the equation of circle circumscribing the given triangle be

x^{2}+y^{2}+2gx+2fy+c=0

36+64+12g+16f+c=0⇒12g+16f+c=-100

16+36+8g+12f+c=0⇒8g+12f+c=-52

0+0+0+0+c=0⇒c=0

3g+4f=-25

2g+3f=-13

On solving we get g=-23 and f=11

The equation of circle is x^{2}+y^{2}-46x+22y=0

Question:43

Answer:

Let equation of ellipse be

a=3 and b=2

b^{2}=a^{2}(1-e^{2})

e^{2}=1-4/9=5/9

From the definition of the ellipse for any point P on the ellipse we have

SP+S'P=2a

Length of endless string=SP+S'P+SS'=2a+2ae=

Question:44

The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is_________

Answer:

be=1

Length of minor axis 2a=1

a=1/2

a^{2}=b^{2}(1-e^{2})

1/4=b^{2}-b^{2}e^{2}

b^{2}=1/4+1=5/4

So, the equation of ellipse is

Question:45

Answer:

Let any point on parabola be P(x,y)

Length of perpendicular from S on the directrix=SP

4x^{2}+y^{2}+4x+32y+16=0

Question:46

Answer:

Let the equation of parabola be

b=6

e=5/3

a^{2}=b^{2}(e^{2}-1)

a^{2}=64

Question:47

The area of the circle centred at (1, 2) and passing through (4, 6) is

A. 5π

B. 10π

C. 25π

D. none of these

Answer:

Centre of the circle is C(1,2)

Radius=

Area of circle= π(5)^{2}=25π square units

Question:48

Equation of a circle which passes through (3, 6) and touches the axes is

A. x^{2} + y^{2 }+ 6x + 6y + 3 = 0

B. x^{2} + y^{2} – 6x – 6y – 9 = 0

C. x^{2 }+ y^{2} – 6x – 6y + 9 = 0

D. none of these

Answer:

Given that the circle touches both axes

Therefore, equation of the circle is (x-a)^{2}+(y-a)^{2}=a^{2}

circle passes through (3,6)

(3-a)^{2}+(6-a)^{2}=a^{2}

a^{2}-18a+45=0

(a-3)(a-15)=0

a = 3 or a =15

For a=3 equation of circle

(x-3)^{2}+(y-3)^{2}=9

x^{2}+y^{2}-6x-6y+9=0

Question:49

Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is

A. x^{2} + y^{2} + 13y = 0

B. 3x^{2} + 3y^{2} + 13x + 3 = 0

C. 6x^{2} + 6y^{2} – 13x = 0

D. x^{2} + y^{2} +13x +3 = 0

Answer:

Centre of the circle lies on the y axis

So, let the centre be C(0,k)

Circle passes through O(0,0)and A(2,3)

OC^{2}=AC^{2}

k^{2}=(2-0)^{2}+(3-k)^{2}

k=13/6

(x-0)^{2}+(y-13/6)^{2}=(13/6)^{2}

3x^{2}+3y^{2}-13y=0

Answer- None

Question:50

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is

A. x^{2} + y^{2} = 9a^{2}

B. x^{2} + y^{2} = 16a^{2}

C.x^{2} + y^{2} = 4a^{2}

D. x^{2} + y^{2} = a^{2}

Answer:

Option (C)

median length=3a

Radius of circle=2/3*median length=2a

Equation of circle

x^{2}+y^{2}=4a^{2}

Question:51

If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is

A. x^{2} = –12y

B. x^{2} = 12y

C. y^{2} = –12x

D. y^{2 }= 12x

Answer:

a Given that, focus of parabola is at S(0,-3)and equation of directrix is y=3

For any point P(x,y)on the parabola, we have

SP=PM

x^{2}+y^{2}+6y+9=y^{2}-6y+9

x^{2}=-12y

Question:52

If the parabola y^{2} = 4ax passes through the point (3, 2), then the length of its latus rectum is

A. 2/3

B. 4/3

C. 1/3

D. 4

Answer:

Option (b) is correct.

Parabola y^{2}=4ax, passes through the point (3,2)

4=4a(3)

a=1/3

Length of latus rectum 4a=4/3

Question:53

If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is

A. y^{2} = 8 (x + 3)

B. x^{2} = 8 (y + 3)

C. y^{2} = – 8 (x + 3)

D. y^{2} = 8 ( x + 5)

Answer:

Given: Vertex =(–3, 0) and the directrix x = -5(M)

So focus S (-1,0)

For any point of parabola P (x,y) we have

SP=PM

x^{2}+y^{2}+2x+1=x^{2}+10x+25

y^{2}=8x+24

Question:54

The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity 1/2 is

A. 7x^{2} + 2xy + 7y^{2} – 10x + 10y + 7 = 0

B. 7x^{2 }+ 2xy + 7y^{2} + 7 = 0

C. 7x^{2 }+ 2xy + 7y^{2} + 10x – 10y – 7 = 0

D. none

Answer:

Option (a) is correct.

Given that focus of the ellipse is S(1,-1)and the equation of directrix is x-y-3=0

Also, e=1/2

From definition of ellipse, for any point P (x,y) on the ellipse, we have SP=ePM, where

M is foot of the perpendicular from point P to the directrix.

7x^{2}+7y^{2}+2xy-10x+10y+7=0

Question:55

** **The length of the latus rectum of the ellipse 3x^{2 }+ y^{2} = 12 is

A. 4

B. 3

C. 8

D. 12

Answer:

Option (d) is correct.

Given ellipse is 3x^{2}+y^{2}=12

a^{2}=4

a=2

b^{2}=12

b=

length of latus rectum=

Question:56

If e is the eccentricity of the ellipse , then

A. b^{2} = a^{2} (1 – e^{2})

B. a^{2} = b^{2} (1 – e^{2})

C. a^{2} = b^{2} (e^{2} – 1)

D. b^{2} = a^{2} (e^{2} – 1)

Answer:

Option (b) is correct.

Given that

We know that a^{2}=b^{2}(1-e^{2})

Question:57

The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is

A. 4/3

B.

C.

D. none of these

Answer:

Let the equation of the hyperbola be x^{2}/a^{2}-y^{2}/b^{2}=1

length of latus rectum=8

2b^{2}/a=8

b^{2}=4a

Conjugate axis=half of the distance between the foci

2b=ae

b^{2}=a^{2}(e^{2}-1)

=a^{2}(e^{2}-1)

e^{2}=4/3

e=

Question:58

The distance between the foci of a hyperbola is 16 and its eccentricity is . Its equation is

A. x^{2} – y^{2} = 32

B.

C. 2x – 3y^{2} = 7

D. none of these

Answer:

Option (A) is correct.

Let the equation of the hyperbola be

Given that Foci=(±ae,0)

Distance between foci=2ae=16

=32(2-1)=32

Hence, equation is

Question:59

Equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0) is

A.

B.

C.

D. None of these

Answer:

Option (A) is correct.

Let the equation of the hyperbola be

e=3/2

Foci=(±ae),0=(±2,0)

So, after comparing the equations, ae=2

a*3/2=2

a=4/3

b^{2}=a^{2}(e^{2}-1)

b^{2}=(4/3)^{2}((3/2)^{2}-1)

=(16/9)(9/4-1)

=16/9*5/4=20/9

Equation of hyperbola is

=

Hence, is the required equation.

The students aiming at achieving the best possible solutions for NCERT questions can access the study material through NCERT Exemplar Class 11 Maths solutions chapter 11 PDF download that is drafted by professionals for expert guidance.

NCERT Exemplar solutions for Class 11 Maths chapter 11 help students for getting an easy yet effective approach to NCERT problems that are important for their exams and carefully drafted by studying the CBSE guidelines for problem-solving.

- 11.1 Introduction
- 11.2 Sections of a cone
- 11.2.1 Circle, ellipse, parabola and hyperbola
- 11.2.2 Degenerated Conic section
- 11.3 Circle
- 11.4 Parabola
- 11.4.1 Standard equations of parabola
- 11.4.2 Latus Rectum
- 11.5 Ellipse
- 11.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the center of the ellipse
- 11.5.2 Special cases of an ellipse
- 11.5.3 Eccentricity
- 11.5.4 Standard equation of an ellipse
- 11.5.5 Latus Rectum
- 11.6 Hyperbola
- 11.6.1 Eccentricity
- 11.6.2 Standard equation of a hyperbola

With Class 11 Maths NCERT exemplar solutions chapter 11, the students will be introduced to the study of a conic section in a two-dimensional plane through the use of algebra and geometry that has a number of uses in various other fields of science.

- The students interested in pursuing any field or branch of science in future will be guided a lot through NCERT Exemplar Class 11 Maths solutions chapter 11 as it has wide application in different fields including physics for understanding and examination of different laws and every other analysis and derivation along with the subject.
- It also represents great ways to display the relationship between analytical and synthetic thinking in mathematics through the study to coordinate planes.
- The chapter also aids in the introduction of matrices and determinants and their application for further study of curves and multivariable calculus.
- The students will also learn about the geometry of different conic sections, including circle, ellipse, parabola and hyperbola along with the knowledge of equations of varied conic sections.

**NCERT Exemplar Class 11 Maths Solutions Chapter-Wise**

Some of the important topics for students to review from this chapter are:

· The students will learn about the representation of curved figures such as circle, ellipse, hyperbola and parabola in the coordinate plane.

· NCERT Exemplar Class 11 Maths solutions chapter 11 covers standard equations of circle and parabola.

· The students must cover standard equations of parabola and hyperbola in addition to their eccentricity and latus rectrum.

· The students must thoroughly practice the solutions and examples of NCERT for examination.

**Check Chapter-Wise NCERT Solutions of Book**

Chapter-1 | |

Chapter-2 | |

Chapter-3 | |

Chapter-4 | |

Chapter-5 | |

Chapter-6 | |

Chapter-7 | |

Chapter-8 | |

Chapter-9 | |

Chapter-10 | |

Chapter-11 | Conic Section |

Chapter-12 | |

Chapter-13 | |

Chapter-14 | |

Chapter-15 | |

Chapter-16 |

**Read more NCERT Solution subject wise -**

- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Physics
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology

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Download EBook**Also, read NCERT Notes subject wise -**

- NCERT Notes for Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology

**Also Check NCERT Books and NCERT Syllabus here:**

1. Who has prepared these solutions for NCERT exercises?

NCERT Exemplar Class 11 Maths chapter 11 solutions provided on our website are created by a trained professional of mathematics that provides a better understanding and easier methods through research of various problem-solving methods suited to CBSE.

2. Are these solutions reliable from the exam point of view?

Yes, the NCERT exemplar solutions for Class 11 Maths chapter 11 developed by the experts are very reliable from the perspective of exams and are drafted in a way to satisfy the methods and guidelines of CBSE and NCERT.

3. How are these solutions helpful for competitive exams?

The NCERT Exemplar Class 11 Maths solutions chapter 11 provided here are very well examined and suggest the best possible method for solving any problem related to the concept.

Sep 06, 2024

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