NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

# NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

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 WallahAakash Unacademy

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

## NCERT Exemplar Class 11 Maths Solutions Chapter 11: Exercise-1.3

Question:1

Given that the circle has a radius aand touches both axis. So the centre is (a,a).
(x-a)2+(y-a)2=a2
x2+y2-2ax-2ay+a2=0

Question:2

Show that the point (x, y) given by and lies on a circle for all real values of t such that –1 ≤ t ≤1 where a is any given real numbers.

We have variable point as and
$\\x^{2}+y^{2}=\frac{4a^{2}t^{2}}{\left ( 1+t^{2} \right )^{2}}+\frac{a^{2}\left ( 1+t^{4}-2t^{2} \right )}{\left ( 1+t^{2} \right )^{2}} \\\\=\frac{a^{2}+a^{2}t^{4}+2a^{2}t^{2}}{\left ( 1+t^{2} \right )^2}\\\\=\frac{a^{2}\left ( 1+t^{2} \right )^{2}}{\left ( 1+t^{2} \right )^2}\\\\= a^{2}$

Question:3

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).

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
x2-2x+1+y2-4y+4=4
x2+y2-2x-4y+1=0

Question:5

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.

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=

Question:6

Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0and lies in the third quadrant.

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

a=2 or a=-4/3
a=2
Equation of required circle
(x+2)2+(y+2)2=22
x2+y2+4x+4y+4=0

Question:7

If one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.

Given equation of the circle is
x2+y2-4x-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 (x1, y1)
and
x1=1 and y1=2

Question:8

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

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= $\sqrt{\left (4-1 \right )^{2}+\left (2-(-2 \right )^{2}}=\sqrt{9+16}=5$
So, equation of the required circle is:
(x-1)2+(y+2)2=52
x2+y2-2x+4y-20=0

Question:9

If the line touches the circle x2 + y2 = 16, then find the value of k.

Given line is and the circle is
x2+y2=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

Find the equation of a circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and has double of its area.

Given equation of the circle is :
x2+y2-6x+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).
Area of circle=
r2=60
$r = \sqrt{60}$
Equation of required circle (x-3)2+(y+6)2=60
x2+y2-6x+12y-15=0

Question:11

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

Consider the equation of the ellipse as
It is given that length of latus rectum = Half of minor axis

a=2b
b2=a2(1-e2)
b2=4b2(1-e2)
1-e2=1/4
e2=3/4

Question:12

Given the ellipse with equation 9x2 + 25y2 = 225, find the eccentricity and foci.

Given equation of ellipse
9x2+25y2=225

a=5, b=3
b2=a2(1-e2)
9=25(1-e2)

$\text { Foci } \equiv(\pm a e, 0) \equiv(\pm 5 \times(4 / 5), 0) \equiv(\pm 4,0)$

Question:13

If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.

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).

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

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 y2 = 8x whose focal distance is 4.

Given parabola is y2=8x
On comparing this parabola to the y2=4ax
we get a=2
Focal distance=Distance of any point on parabola from the focus SP=

|x1+2|=4
x1=2,-6
But x≠ -6
So for x=2

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

Question:18

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

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

x2+y2-4y+4=y2-12y+36
x2+8y=32

Question:19

If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.

Given that line y=mx+1 is tangent to the parabola y2=4x
Solving line with parabola we have
(mx+1)2=4x
m2x2+2mx+1=4x
m2x2+x(2m-4)+1=0
Since the line touches the parabola, above equation must have equal roots.
Discriminant D=0
(2m-4)2-4m2=0
4m2-16m+16-4m2=0
16m=16
m=1

Question:20

Let the equation of the hyperbola be
Distance between foci=2ae=16

We know that

x2-y2=32

Question:21

Find the eccentricity of the hyperbola 9y2 – 4x2 = 36

We have the hyperbola 9y2-4x2=36

a2=9
b2=4
a2=b2(e2-1)

Question:22

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

Let the equation of the hyperbola be
ae=2

b2=a2(e2-1)

Question:23

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

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
πr2=154
r2=154*7/22=49
r=7
(x-1)2+(y+1)2=49
x2+y2-2x+2y-47=0

Question:24

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

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
AC2=BC2
(h-2)2+(4h-3-3)2=(h-4)2+(4h-3-5)2
h2-4h+4+16h2-48h+36=h2-8h+16+16h2-64h+64
20h=40
h=2
k=4h-3=5
Therefore equation of the circle is : (x-2)2+(y-5)2=4
x2+y2-4x-10y+25=0

Question:25

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

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=
OB2=OP2+PB2
OB2=29+9=38
OB=
Equation of the circle is (x-3)2+(y+1)2=38
x2+y2-6x+2y=28

Question:26

Find the equation of a circle of radius 5 which is touching another circle x2 + y2 – 2x – 4y – 20 = 0 at (5, 5).

Given circle is x2+y2-2x-4y=20
(x-1)2+(y-2)2=52
Centre of this circle is C1(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 C2(h,k).
Since the radius of the given circle and the required circle is same, point P is mid point of C1C2.
and
h=9 and k=8
So the equation of the required circle is (x-9)2+(y-8)2=25
x2+y2-18x-16y+120=0

Question:27

Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x – 1.

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=32
2h2-22h+56=0
h2-11h+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⇒x2+y2-8x-6y+16=0
If the centre is C(7,6)⇒(x-7)2+(y-6)2=9⇒x2+y2-14x-12y+76=0

Question:28

Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0

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 x2=(x-6)2+y2
y2-12x+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

y2-12y+36=x2+y2-4y+4
x2=32-8y
iii) Given that focus is at (-1,-2)
and directrix x-2y+3=0

x2+2x+1+y2+4y+4=1/5[x2+4y2+9+6x-4xy-12y]
4x2+4xy+y2+4x+32y+16=0

Question:29

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Let the coordinates of the variable point be (x,y)
Then according to the question
x2-6x+9+y2=

Again squaring both sides, we get
x2-36x+324=4(x2-18x+81+y2)
3x2+4y2-36x=0 which is an ellipse.

Question:30

Find the equation of the set of all points whose distance from (0, 4) are 2/3 of their distance from the line y = 9.

Let the point be P (x,y)
According to the question

9(x2+y2-8y+16)=4(y2-18y+81)
9x2+5y2=180

Question:31

Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represent a hyperbola.

Let the point be P (x,y)
According to the question
Distance of P from (-4,0) -Distance of P from (4,0)=2
$\sqrt{\left ( x+4 \right )^{2}+y^{2}}=2+\sqrt{\left ( x-4 \right )^{2}+y^{2}}$
Squaring both the sides
$x^{2}+8x+16+y^{2}=4+x^{2}-8x+16+y^{2}+4\sqrt{\left ( x-4 \right )^{2}+y^{2}}$
$4x-1=\sqrt{\left ( x-4 \right )^{2}+y^{2}}$
16x2-8x+1=x2+16-8x+y2
15x2-y2=15
Which is an equation of a hyperbola.

Question:32

Find the equation of the hyperbola with

(i) Vertices (± 5, 0), foci (± 7, 0)
(ii) Vertices (0, ± 7), e =
(iii) Foci , passing through (2, 3)

a) Given that a=5 and ae=7
e=7/5
b2=a2(e2-1)=25(49/25-1)=24
So, the equation of hyperbola is

b) b=7,e=4/3

c) Given that foci=(0,±10 ) $be= \sqrt{10}$
a2=b2(e2-1)
a2=b2e2-b2
a2=10-b2
Since, hyperbola passes through the point (2,3)
4b2-9(10-b2)=-b2(10-b2)
b4-23b2+90=0
b4-18b2-5b2+90=0
(b2-18)(b2-5)=0
b2=5
a2+b2=10⇒a2=5

y2-x2=5

Question:33

The line x + 3y = 0 is a diameter of the circle x2 + y2 + 6x + 2y = 0.

False
Given equation of the circle is x2+y2+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:34

The shortest distance from the point (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = 0 is equal to 5.

False
$\text { Centre } \equiv C(7,5) \\$
$Radius =\sqrt{49+25+151}=\sqrt{225}=15$
$\\Distance between the point P(2,-7) and centre =\sqrt{(2-7)^{2}+(-7-5)^{2}}$=$\sqrt{169}=13$

Question:35

If the line lx + my = 1 is a tangent to the circle x2 + y2 = a2, then the point (l, m) lies on a circle.

True
Given circle is x2+y2=a2
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

$l^2+m^{2}=\frac{1}{a^{2}}$
The locus of (l,m) is . which is an equation of a circle.

False
Given circle is x2+y2-2x+6y+1=0 or (x-1)2+(y+3)2=32
Centre is C(1,-3) and radius is 3.
Distance of point P (1,2) from centre is 5.
So, point P lies outside the circle.

Question:37

The line lx + my + n = 0 will touch the parabola y2 = 4ax if ln = am2.

True
Give line lx+my+n=0
and parabola y2=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

am2=nl

Question:38

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).

True
Given line is 2x+3y=12
and the ellipse 4x2+9y2=72
Solving line and ellipse we get (12-3y)2+9y2=72
(4-y)2+y2=8
2y2-8y+8=0
y2-4y+4=0
(y-2)2=0
y=2;x=3

Question:40

The locus of the point of intersection of lines and for different value of k is a hyperbola whose eccentricity is 2.

True
Given equation of lines are

and

3x2-y2=48
which is the equation of hyperbola
a2=16 and b2=48
e2=1+48/16=4
e=2

Question:41

The equation of the circle having centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ______.

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

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, 2y = 3x is __________ .

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
x2+y2+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 x2+y2-46x+22y=0

Question:43

An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____________.

Let equation of ellipse be
a=3 and b=2
b2=a2(1-e2)
e2=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_________

be=1
Length of minor axis 2a=1
a=1/2
a2=b2(1-e2)
1/4=b2-b2e2
b2=1/4+1=5/4

So, the equation of ellipse is

Question:45

The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ____________ .

Let any point on parabola be P(x,y)
Length of perpendicular from S on the directrix=SP

4x2+y2+4x+32y+16=0

Question:46

The equation of the hyperbola with vertices at (0, ± 6) and eccentricity 5/3 is ________ and its foci are __________ .

Let the equation of parabola be
b=6
e=5/3
a2=b2(e2-1)
a2=64

$\text { foci }=(0, \pm b e) \equiv\left(0, \pm \frac{5}{3} \times 6\right)=(0,\pm 10)$

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

Centre of the circle is C(1,2)
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. x2 + y2 + 6x + 6y + 3 = 0
B. x2 + y2 – 6x – 6y – 9 = 0
C. x2 + y2 – 6x – 6y + 9 = 0
D. none of these

Given that the circle touches both axes
Therefore, equation of the circle is (x-a)2+(y-a)2=a2
circle passes through (3,6)
(3-a)2+(6-a)2=a2
a2-18a+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
x2+y2-6x-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. x2 + y2 + 13y = 0
B. 3x2 + 3y2 + 13x + 3 = 0
C. 6x2 + 6y2 – 13x = 0
D. x2 + y2 +13x +3 = 0

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)
OC2=AC2
k2=(2-0)2+(3-k)2
k=13/6
(x-0)2+(y-13/6)2=(13/6)2
3x2+3y2-13y=0

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. x2 + y2 = 9a2
B. x2 + y2 = 16a2
C.x2 + y2 = 4a2
D. x2 + y2 = a2

Option (C)
median length=3a
Equation of circle
x2+y2=4a2

Question:51

If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
A. x2 = –12y
B. x2 = 12y
C. y2 = –12x
D. y2 = 12x

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

x2+y2+6y+9=y2-6y+9
x2=-12y

Question:52

Option (b) is correct.
Parabola y2=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. y2 = 8 (x + 3)
B. x2 = 8 (y + 3)
C. y2 = – 8 (x + 3)
D. y2 = 8 ( x + 5)

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

x2+y2+2x+1=x2+10x+25
y2=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. 7x2 + 2xy + 7y2 – 10x + 10y + 7 = 0
B. 7x2 + 2xy + 7y2 + 7 = 0
C. 7x2 + 2xy + 7y2 + 10x – 10y – 7 = 0
D. none

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.

7x2+7y2+2xy-10x+10y+7=0

Question:55

The length of the latus rectum of the ellipse 3x2 + y2 = 12 is
A. 4
B. 3
C. 8
D. 12

Option (d) is correct.
Given ellipse is 3x2+y2=12

a2=4
a=2
b2=12
b=
length of latus rectum=

Question:56

If e is the eccentricity of the ellipse , then
A. b2 = a2 (1 – e2)
B. a2 = b2 (1 – e2)
C. a2 = b2 (e2 – 1)
D. b2 = a2 (e2 – 1)

Option (b) is correct.
Given that
We know that a2=b2(1-e2)

Question:57

Let the equation of the hyperbola be x2/a2-y2/b2=1
length of latus rectum=8
2b2/a=8
b2=4a
Conjugate axis=half of the distance between the foci
2b=ae
b2=a2(e2-1)
=a2(e2-1)
e2=4/3
e=

Question:58

The distance between the foci of a hyperbola is 16 and its eccentricity is . Its equation is
A. x2 – y2 = 32
B.
C. 2x – 3y2 = 7
D. none of these

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

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
b2=a2(e2-1)
b2=(4/3)2((3/2)2-1)
=(16/9)(9/4-1)
=16/9*5/4=20/9
Equation of hyperbola is
=

Hence, $\frac{x^{2}}{16}-\frac{y^{2}}{{20}}=\frac{1}{9}$ is the required equation.

## More About NCERT Exemplar Class 11 Maths Solutions Chapter 11

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.

### Topics and Subtopics in NCERT Exemplar Class 11 Maths Solutions Chapter 11

• 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

## What will the students learn in NCERT Exemplar Class 11 Maths Solutions Chapter 11?

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

 Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutations and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight lines Chapter 12 Introduction to Three Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

### Important topics to cover from NCERT Exemplar Class 11 Maths Solutions Chapter 11

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 Sets Chapter-2 Relations and Functions Chapter-3 Trigonometric Functions Chapter-4 Principle of Mathematical Induction Chapter-5 Complex Numbers and Quadratic equations Chapter-6 Linear Inequalities Chapter-7 Permutation and Combinations Chapter-8 Binomial Theorem Chapter-9 Sequences and Series Chapter-10 Straight Lines Chapter-11 Conic Section Chapter-12 Introduction to Three Dimensional Geometry Chapter-13 Limits and Derivatives Chapter-14 Mathematical Reasoning Chapter-15 Statistics Chapter-16 Probability

### NCERT Exemplar Class 11 Solutions

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