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NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Edited By Komal Miglani | Updated on Mar 30, 2025 07:44 AM IST

Have you ever noticed the shape of satellite dishes, car headlights, or even the paths of planets? These shapes come from conic sections, a key topic in geometry. This chapter, conic sections, is a continuation of the last chapter of straight lines, and here we will discuss mainly four other curves: Circle, Hyperbola, Parabola, and Ellipse. “Conic” means “Cone” and “Sections” means “Intersection”. These curves are called “Conic sections” because they can be formed by the intersection of a plane with a double-napped right circular cone. Now, strengthening the basic concepts is necessary, and the NCERT Solutions will help to command the concepts.

This Story also Contains
  1. Conic Sections Class 11 Questions And Answers PDF Free Download
  2. Conic Sections Class 11 Solutions: Important Formulae
  3. Conic Sections Class 11 NCERT Solutions (Exercise)
  4. Importance of solving NCERT Questions for Class 11 Chapter 10 Conic Sections
  5. NCERT Solutions For Class 11: Subject Wise
  6. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

This article will explain all the Conic Sections class 11 questions and answers based on the latest NCERT 2025-26 syllabus in detail. Conic sections NCERT solutions are not only useful for CBSE board exams but also for competitive exams like JEE Mains, JEE Advanced, VITEEE, and BITSAT. Learnings from this chapter can be used in real-life applications like Wheels, satellite dishes, headlights, GPS, planetary orbits, MRI machines and many more. For a quick revision, students can use the PDFs of class 11 Maths chapter 10 NCERT solutions. Conic Section Class 11 Maths Chapter Notes and NCERT Exemplar Solutions For Class 11 Maths Chapter Conic Sections can also be used for mastering this chapter and deeper knowledge gain purposes.

Conic Sections Class 11 Questions And Answers PDF Free Download

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Conic Sections Class 11 Solutions: Important Formulae

Circle

Description

Equation/Formulae

Equation of a circle

(x - h)2 + (y - k)2 = r2

General equation of a circle

x2 + y2 + 2gx + 2fy + c = 0

Centre of the circle

Centre: (-g, -f)

Radius of the circle

Radius (r) = g2+f2c

Parametric equation of a circle

x = r cos(θ), y = r sin(θ)

Parametric equation of a circle (centre at h, k)

x = h + r cos(θ), y = k + r sin(θ)


Parabola

Description

Equations/Forms

Equation forms of parabola

y2 = 4ax, y2 = -4ax, x2 = 4ay, x2 = -4ay

Axis of the parabola

y = 0 (for first two forms), x = 0 (for last two forms)

Directrix of the parabola

x = -a (1st form), x = a (2nd form), y = -a (3rd form), y = a (4th form)

Vertex of the parabola

(0, 0) (for all forms)

Focus of the Parabola

(a, 0) (1st form), (-a, 0) (2nd form), (0, a) (3rd form), (0, -a) (4th form)

Length of the latus rectum

4a (for all forms)

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Ellipse

Description

Equation/Forms

Equation forms of ellipse

x2a2+y2b2=1(a>b
And
x2b2+y2a2=1(a>b)

Major Axis

y = 0 (1st form), x = 0 (2nd form)

Length of Major Axis

2a (for both forms)

Minor Axis

x = 0 (1st form), y = 0 (2nd form)

Length of Minor Axis

2b (for both forms)

Directrix of the ellipse

x = ±ae (1st form), y = ±ae (2nd form)

Vertex of the ellipse

(±a, 0) (1st form), (0, ±a) (2nd form)

Focus of the ellipse

(±ae, 0) (1st form), (0, ±ae) (2nd form)

Length of the latus rectum

2b2a (for both forms)

Eccentricity (e)

a2+b2a2(for both forms)


Hyperbola

Description

Equations/Forms

Equation forms of the hyperbola

x2a2+y2b2=1
And
x2b2+y2a2=1

Coordinates of centre

(0, 0) (for both forms)

Coordinates of vertices

(±a, 0) (for both forms)

Coordinates of foci

(±ae, 0) (for both forms)

Length of Conjugate axis

2b (for both forms)

Length of Transverse axis

2a (for both forms)

Equation of Conjugate axis

x = 0 (for both forms)

Equation of Transverse axis

y = 0 (for both forms)

Equation of Directrix

x = ±ae (for both forms)

Eccentricity (e)

(a2+b2)a2(for both forms)

Length of the latus rectum

2b2a (for both forms)

Conic Sections Class 11 NCERT Solutions (Exercise)

NCERT Conic Sections Class 11 Solutions: Exercise: 11.1
Page Number: 181
Total Questions: 15

Question:1 Find the equation of the circle with centre (0,2) and radius 2.

Answer:
As we know, the equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here, (h,k)=(0,2) and r=2

So the equation of the circle is:

(x0)2+(y2)2=22

x2+y24y+4=4

x2+y24y=0

Question:2 Find the equation of the circle with centre (–2,3) and radius 4.

Answer:
As we know,
The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here (h,k)=(2,3) and r=4

So the equation of the circle is:

(x(2))2+(y3)2=42

x2+4x+4+y26y+9=16

x2+y2+4x6y3=0

Question:3 Find the equation of the circle with centre (12,14) and radius 112.

Answer:
As we know, the equation of the circle with centre ( h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here, (h,k)=(12,14) and r=112

So the equation of the circle is:

(x12)2+(y14)2=(112)2

x2x+14+y212y+116=1144

x2+y2x12y1136=0

36x2+36y236x18y11=0

Question:4 Find the equation of the circle with centre (1,1) and radius 2.

Answer:

As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here (h,k)=(1,1) and r=2

So the equation of the circle is:

(x1)2+(y1)2=(2)2

x22x+1+y22y+1=2

x2+y22x2y=0

Question:5 Find the equation of the circle with centre (a,b) and radius a2b2.

Answer:
As we know,

The equation of the circle with centre ( h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here (h,k)=(a,b) and r=a2b2

So the equation of the circle is:

(x(a))2+(y(b))2=(a2b2)2

x2+2ax+a2+y2+2by+b2=a2b2

x2+y2+2ax+2by+2b2=0

Question:6 Find the centre and radius of the circles: (x+5)2+(y3)2=36

Answer:
As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here, (x+5)2+(y3)2=36

It can also be written in the form: (x(5))2+(y3)2=62
After comparing, we can see that,
r=6

Hence, the radius of the circle is 6.

Question:7 Find the centre and radius of the circles: x2+y24x8y45=0

Answer:
As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here x2+y24x8y45=0

It can also be written in the form: (x2)2+(y4)2=(65)2

After comparing, we can see that,

r=65

Hence, the radius of the circle is 65.

Question:8 Find the centre and radius of the circles: x2+y28x+10y12=0

Answer:
As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here, x2+y28x+10y12=0

It can also be written in the form: (x4)2+(y(5))2=(53)2

After comparing, we can see that

r=53

Hence, the radius of the circle is 53.

Question:9 Find the centre and radius of the circles: 2x2+2y2x=0

Answer:

As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Here, 2x2+2y2x=0

It can also be written in the form: (x14)2+(y0)2=(14)2

After comparing, we can see that,

r=14

Hence, centre of the circle is the (14,0) and radius of the circle is 14.

Question:10: Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x+y=16.

Answer:

As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

Condition 1: The circle passes through points (4,1) and (6,5).

(4h)2+(1k)2=r2

(6h)2+(5k)2=r2

Here,

(4h)2+(1k)2=(6h)2+(5k)2

(4h)2(6h)2+(1k)2(5k)2=0

(2)(102h)+(4)(62k)=0

20+4h24+8k=0

4h+8k=44

Condition 2: Centre is on the line 4x+y=16.

So, 4h+k=16

From condition 1 and condition 2, we get,

h=3 and k=4

Substituting the values of h and k in condition 1, we get,

(43)2+(14)2=r2

1+9=r2

r=10

Final Equation of the circle is:

(x3)2+(y4)2=(10)2

x26x+9+y28y+16=10

x2+y26x8y+15=0

Question:11: Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x3y11=0.

Answer:

As we know,

The equation of the circle with centre ( h, k) and radius r is given by ;

(xh)2+(yk)2=r2

Condition 1: The circle passes through points (2,3) and (–1,1).

(2h)2+(3k)2=r2

(1h)2+(1k)2=r2

Here,

(2h)2+(3k)2=(1h)2+(1k)2

(2h)2(1h)2+(3k)2(1k)2=0

(3)(12h)+(2)(42k)=0

36h+84k=0

6h+4k=11

Condition 2: The centre is on the line x3y11=0

So, h3k=11

From condition 1 and condition 2, we get,

h=72 and k=52

Substituting the values of h and k in condition 1, we get,

(272)2+(3+52)2=r2

94+1214=r2

r2=1304

Final Equation of the circle is:

(x72)2+(y+52)2=1304

x27x+494+y2+5y+254=1304

x2+y27x+5y564=0

x2+y27x+5y14=0

Question:12: Find the equation of the circle with radius 5, whose centre lies on the x-axis and passes through the point (2,3).

Answer:

As we know,

The equation of the circle with centre (h, k) and radius r is given by:

(xh)2+(yk)2=r2

So let the circle be,

(xh)2+(yk)2=r2

Since its radius is 5 and its centre lies on the x-axis,

(xh)2+(y0)2=52

And since it passes through the point (2,3).

(2h)2+(30)2=52

(2h)2=259

(2h)2=16

(2h)=4 or, (2h)=4

h=2 or, h=6

When h=2, the equation of the circle is:

(x(2))2+(y0)2=52

x2+4x+4+y2=25

x2+y2+4x21=0

When h=6, the equation of the circle is:

(x6)2+(y0)2=52

x212x+36+y2=25

x2+y212x+11=0

Question:13: Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Answer:

Let the equation of circle be (xh)2+(yk)2=r2
Now, since this circle passes through (0, 0),

(0h)2+(0k)2=r2

h2+k2=r2

Now, this circle makes an intercept of a and b on the coordinate axes. it means the circle passes through the points (a, 0) and (0, b).

So,

(ah)2+(0k)2=r2

a22ah+h2+k2=r2

a22ah=0

a(a2h)=0

a=0 or a2h=0
Since a0, So, a2h=0

h=a2

Similarly,

(0h)2+(bk)2=r2

h2+b22bk+k2=r2

b22bk=0

b(b2k)=0
Since b is not equal to zero.

k=b2

So the final equation of the Circle is:

(xa2)2+(yb2)2=(a2)2+(b2)2

x2ax+a24+y2bx+b24=a24+b24

x2+y2axbx=0

Question:14: Find the equation of a circle with a centre (2,2) and passes through the point (4,5).

Answer:

Let the equation of the circle be:

(xh)2+(yk)2=r2

Now, since the centre of the circle is (2,2), our equation becomes:

(x2)2+(y2)2=r2

Now, since this passes through the point (4,5),

(42)2+(52)2=r2

4+9=r2

r2=13

Hence, The Final equation of the circle becomes:

(x2)2+(y2)2=13

x24x+4+y24y+4=13

x2+y24x4y5=0

Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle x2+y2=25?

Answer:

Given, a circle x2+y2=25

As we can see, the center of the circle is (0, 0).

Now the distance(d) between (0,0) and (–2.5, 3.5) is:

d=(2.50)2+(3.50)2

d=6.25+12.25

d=18.54.3<5

Since the distance between the given point and the centre of the circle is less than the radius of the circle, the point lies inside the circle.

NCERT Conic Sections Class 11 Solutions: Exercise: 11.2
Page Number: 186-187
Total Questions: 12

Question:1: Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: y2=12x

Answer:

Given a parabola with equation y2=12x

This is a parabola of the form y2=4ax, which opens towards the right.
So, by comparing the given parabola equation with the standard equation, we get,

4a=12
a=3

Hence,

Coordinates of the focus :

(a,0)=(3,0)

Axis of the parabola:

It can be seen that the axis of this parabola is the X-axis.

The equation of the directrix:
x=a
x=3
x+3=0
The length of the latus rectum:
4a=4(3)=12

Question:2: Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: x2=6y

Answer:

Given a parabola with equation x2=6y

This is a parabola of the form x2=4ay, which opens upward.

So, by comparing the given parabola equation with the standard equation, we get,

4a=6

a=32

Hence,

Coordinates of the focus :

(0,a)=(0,32)

Axis of the parabola:

It can be seen that the axis of this parabola is the Y-axis.

The equation of the directrix:

y=a
y=32
y+32=0

The length of the latus rectum:

4a=4(32)=6

Question:3 Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: y2=8x

Answer:

Given a parabola with equation y2=8x

This is a parabola of the form y2=4ax, which opens towards the left.

So, by comparing the given parabola equation with the standard equation, we get,

4a=8

a=2

Hence,

Coordinates of the focus :

(a,0)=(2,0)

Axis of the parabola:

It can be seen that the axis of this parabola is the X-axis.

The equation of the directrix:

x=a
x=2
x2=0

The length of the latus rectum:

4a=4(2)=8

Question:4: Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: x2=16y

Answer:

Given a parabola with equation x2=16y

This is a parabola of the form x2=4ay, which opens downward.

So, by comparing the given parabola equation with the standard equation, we get,

4a=16

a=4

Hence,

Coordinates of the focus :

(0,a)=(0,4)

Axis of the parabola:

It can be seen that the axis of this parabola is the Y-axis.

The equation of the directrix

y=a
y=4
y4=0

The length of the latus rectum:

4a=4×4=16

Question:5: Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: y2=10x

Answer:

Given a parabola with equation y2=10x

This is a parabola of the form y2=4ax, which opens towards the right.

So, by comparing the given parabola equation with the standard equation, we get,

4a=10

a=104=52

Hence,

Coordinates of the focus :

(a,0)=(52,0)

Axis of the parabola:

It can be seen that the axis of this parabola is the X-axis.

The equation of the directrix:

x=a
x=52
x+52=0
2x+5=0

The length of the latus rectum:

4a=4×(52)=10

Question:6 Find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum: x2=9y

Answer:

Given a parabola with an equation x2=9y

This is a parabola of the form x2=4ay, which opens downward.

So, by comparing the given parabola equation with the standard equation, we get,

4a=9

a=94

Hence,

Coordinates of the focus :

(0,a)=(0,94)

Axis of the parabola:

It can be seen that the axis of this parabola is the Y-axis.

The equation of the directrix:

y=a
y=94
y94=0

The length of the latus rectum:

4a=4(94)=9

Question:7 Find the equation of the parabola that satisfies the given conditions: Focus (6,0); directrix x=6

Answer:

Given, in a parabola,

Focus : (6,0) And Directrix : x=6

Here,

Focus is of the form (a, 0), which means it lies on the X-axis.
Directrix is of the form x=a, which means it lies left to the Y-Axis.

These are the conditions when the standard equation of a parabola is. y2=4ax

Hence, the Equation of the Parabola is y2=4ax

Here, it can be seen that: a=6

Hence, the Equation of the Parabola is:

y2=4ax

y2=4(6)x

y2=24x .

Question:8 Find the equation of the parabola that satisfies the given conditions: Focus (0,–3); directrix y=3

Answer:

Given, in a parabola,

Focus : Focus (0,–3); directrix y=3

Here,

Focus is of the form (0, -a), which means it lies on the Y-axis.
Directrix is of the form y=a, which means it lies above the X-axis.

These are the conditions when the standard equation of a parabola is x2=4ay. Hence, the Equation of the Parabola is

x2=4ay

Here, it can be seen that:

a=3

Hence, the Equation of the Parabola is:

x2=4ay
x2=4(3)y
x2=12y

Question:9 Find the equation of the parabola that satisfies the given conditions: Vertex (0,0); focus (3,0)

Answer:

Given,

Vertex (0,0) and focus (3,0)

As the vertex of the parabola is (0,0) and the focus lies on the positive X-axis, the parabola will open towards the right. The standard equation of such a parabola is

y2=4ax

Here, it can be seen that a=3

So, the equation of a parabola is

y2=4ax

y2=4(3)x

y2=12x

Question:10 Find the equation of the parabola that satisfies the given conditions: Vertex (0,0); focus (-2,0)

Answer:
Given: Vertex (0,0) and focus (-2,0)

As the vertex of the parabola is (0,0) and the focus lies in the negative X-axis, the parabola will open towards the left, and the standard equation of such a parabola is y2=4ax

Here, it can be seen that a=2

So, the equation of a parabola is:

y2=4ax

y2=4(2)x

y2=8x

Question:11: Find the equation of the parabola that satisfies the given conditions: Vertex (0,0) passing through (2,3) and axis is along the x-axis.

Answer:

Given

The Vertex of the parabola is (0,0).

The parabola is passing through (2,3), and the axis is along the x-axis; it will open towards the right. and the standard equation of such a parabola is

y2=4ax

Now, since it passes through (2,3)

32=4a(2)

9=8a

a=98

So the Equation of the Parabola is:

y2=4(98)x

y2=(92)x

2y2=9x

Question:12: Find the equation of the parabola that satisfies the given conditions: Vertex (0,0), passing through (5,2) and symmetric with respect to the y-axis.

Answer:

Given a parabola with Vertex (0,0), passing through (5,2) and symmetric with respect to the y-axis.

Since the parabola is symmetric with respect to the Y-axis, its axis will be the Y-axis. and since it passes through the point (5,2), it must go through the first quadrant.

So the standard equation of such a parabola is

x2=4ay

Now, since this parabola is passing through (5,2)

52=4a(2)

25=8a

a=258

Hence, the equation of the parabola is:

x2=4(258)y

x2=(252)y

2x2=25y


NCERT Conic Sections Class 11 Solutions: Exercise: 11.3
Page Number: 195
Total Questions: 20

Question:1 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: x236+y216=1

Answer:

The equation of the ellipse x236+y216=1

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

x2a2+y2b2=1

We get,

a=6 and b=4

So,

c=a2b2=6242=20=25

Hence,

Coordinates of the foci:

(c,0) and (c,0)=(25,0) and (25,0)

The vertices:

(a,0) and (a,0)=(6,0) and (6,0)

The length of the major axis:

2a=2(6)=12

The length of the minor axis:

2b=2(4)=8

The eccentricity :

e=ca=256=53

The length of the latus rectum:

2b2a=2(4)26=326=163

Question:2 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: x24+y225=1

Answer:

The equation of the ellipse x24+y225=1

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

x2b2+y2a2=1

We get,

a=5 and b=2

So,

c=a2b2=5222=21

Hence,

Coordinates of the foci:

(0,c) and (0,c)=(0,21) and (0,21)

The vertices:

(0,a) and (0,a)=(0,5) and (0,5)

The length of the major axis:

2a=2(5)=10

The length of the minor axis:

2b=2(2)=4

The eccentricity :

e=ca=216

The length of the latus rectum:

2b2a=2(2)25=85

Question:3 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: x216+y29=1

Answer:

The equation of the ellipse x216+y29=1

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

x2a2+y2b2=1

We get,

a=4 and b=3

So,

c=a2b2=4232=7

Hence, the Coordinates of the foci:

(c,0) and (c,0)=(7,0) and (7,0)

The vertices:

(a,0) and (a,0)=(4,0) and (4,0)

The length of the major axis:

2a=2(4)=8

The length of the minor axis:

2b=2(3)=6

The eccentricity :

e=ca=74

The length of the latus rectum:

2b2a=2(3)24=184=92

Question:4 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: x225+y2100=1

Answer:

The equation of the ellipse x225+y2100=1

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

x2b2+y2a2=1

We get,

a=10 and b=5

So,

c=a2b2=10252=75=53

Hence, the Coordinates of the foci:

(0,c) and (0,c)=(0,53) and (0,53)

The vertices:

(0,a) and (0,a)=(0,10) and (0,10)

The length of the major axis:

2a=2(10)=20

The length of the minor axis:

2b=2(5)=10

The eccentricity :

e=ca=5310=32

The length of the latus rectum:

2b2a=2(5)210=5010=5

Question:5 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: x249+y236=1

Answer:

The equation of ellipse x249+y236=1x272+y262=1

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

x2a2+y2b2=1

We get,

a=7 and b=6 .

So,

c=a2b2=7262=13

Hence, the Coordinates of the foci:

(c,0)and(c,0)=(13,0)and(13,0)

The vertices:

(a,0)and(a,0)=(7,0)and(7,0)

The length of the major axis:

2a=2(7)=14

The length of the minor axis:

2b=2(6)=12

The eccentricity :

e=ca=137

The length of the latus rectum:

2b2a=2(6)27=727

Question:6 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: x2100+y2400=1

Answer:

The equation of the ellipse x2100+y2400=1x2102+y2202=1

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

x2b2+y2a2=1

We get,

a=20 and b=10 .

So,

c=a2b2=202102

c=300=103

Hence,

Coordinates of the foci:

(0,c)and(0,c)=(0,103)and(0,103)

The vertices:

(0,a)and(0,a)=(0,20)and(0,20)

The length of the major axis:

2a=2(20)=40

The length of the minor axis:

2b=2(10)=20

The eccentricity :

e=ca=10320=32

The length of the latus rectum:

2b2a=2(10)220=20020=10

Question:7 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: 36x2+4y2=144

Answer:

The equation of the ellipse 36x2+4y2=144

36144x2+4144y2=1

14x2+136y2=1

x222+y262=1

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

x2b2+y2a2=1

We get,

a=6 and b=2 .

So,

c=a2b2=6222

c=32=42

Hence, the Coordinates of the foci:

(0,c)and(0,c)=(0,42)and(0,42)

The vertices:

(0,a)and(0,a)=(0,6)and(0,6)

The length of the major axis:

2a=2(6)=12

The length of the minor axis:

2b=2(2)=4

The eccentricity :

e=ca=426=223

The length of the latus rectum:

2b2a=2(2)26=86=43

Question:8 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: 16x2+y2=16

Answer:

The equation of the ellipse 16x2+y2=16

16x216+y216=1

x212+y242=1

As we can see from the equation, the major axis is along the Y-axis, and the minor axis is along the X-axis.

On comparing the given equation with the standard equation of such an ellipse, which is

x2b2+y2a2=1

We get,

a=4 and b=1 .

So,

c=a2b2=4212

c=15

Hence, the Coordinates of the foci:

(0,c)and(0,c)=(0,15)and(0,15)

The vertices:

(0,a)and(0,a)=(0,4)and(0,4)

The length of the major axis:

2a=2(4)=8

The length of the minor axis:

2b=2(1)=2

The eccentricity :

e=ca=154

The length of the latus rectum:

2b2a=2(1)24=24=12

Question: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: 4x2+9y2=36

Answer:

The equation of the ellipse 4x2+9y2=36

4x236+9y236=1

x29+y24=1

x232+y222=1

As we can see from the equation, the major axis is along the X-axis, and the minor axis is along the Y-axis.

On comparing the given equation with the standard equation of an ellipse, which is

x2a2+y2b2=1

We get,

a=3 and b=2 .

So,

c=a2b2=3222

c=5

Hence, the Coordinates of the foci:

(c,0)and(c,0)=(5,0)and(5,0)

The vertices:

(a,0)and(a,0)=(3,0)and(3,0)

The length of the major axis:

2a=2(3)=6

The length of the minor axis:

2b=2(2)=4

The eccentricity :

e=ca=53

The length of the latus rectum:

2b2a=2(2)23=83

Question:10: Find the equation for the ellipse that satisfies the given conditions: Vertices (± 5, 0), foci (± 4, 0)

Answer:

Given, in an ellipse,

Vertices (± 5, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

a=5 and c=4

Now, as we know the relation,

a2=b2+c2

b2=a2c2

b=a2c2

b=5242

b=9

b=3

Hence, The Equation of the ellipse will be :

x252+y232=1

x225+y29=1 .

Question:11: Find the equation for the ellipse that satisfies the given conditions: Vertices (0, ± 13), foci (0, ± 5)

Answer:

Given, in an ellipse, Vertices (0, ± 13), foci (0, ± 5)

Here Vertices and focus of the ellipse are on the Y-axis, so the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

x2b2+y2a2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

a=13 and c=5

Now, as we know the relation,

a2=b2+c2

b2=a2c2

b=a2c2

b=13252

b=16925

b=144

b=12

Hence, The Equation of the ellipse will be :

x2122+y2133=1

x2144+y2169=1 .

Question:12: Find the equation for the ellipse that satisfies the given conditions: Vertices (± 6, 0), foci (± 4, 0)

Answer:

Given, in an ellipse, Vertices (± 6, 0), foci (± 4, 0)

Here Vertices and focus of the ellipse are on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

a=6 and c=4

Now, as we know the relation,

a2=b2+c2

b2=a2c2

b=a2c2

b=6242

b=3616

b=20

Hence, The Equation of the ellipse will be :

x262+y2(20)2=1

x236+y220=1 .

Question:13: Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)

Answer:

Given, in an ellipse,

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

Here, the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( ends of the major and minor axis ) with the given one, we get a=3 and b=2

Hence, The Equation of the ellipse will be :

x232+y222=1

x29+y24=1 .

Question:14 Find the equation for the ellipse that satisfies the given conditions: Ends of major axis (0, ± 5 ), ends of minor axis (± 1, 0)

Answer:

Given, in an ellipse,

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

Here, the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

x2b2+y2a2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So on comparing standard parameters( ends of the major and minor axis ) with the given one, we get a=5 and b=1

Hence, The Equation of the ellipse will be :

x212+y2(5)2=1

x21+y25=1 .

Question:15 Find the equation for the ellipse that satisfies the given conditions: Length of major axis 26, foci (± 5, 0)

Answer:

Given, in an ellipse,

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

Here, the focus of the ellipse is on the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So on comparing standard parameters( Length of semimajor axis and foci) with the given one, we get 2a=26a=13 and c=5

Now, as we know the relation,

a2=b2+c2

b2=a2c2

b=a2c2

b=13252

b=144

b=12

Hence, The Equation of the ellipse will be :

x2132+y2122=1

x2169+y2144=1 .

Question:16: Find the equation for the ellipse that satisfies the given conditions: Length of minor axis 16, foci (0, ± 6).

Answer:

Given, in an ellipse,

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

Here, the focus of the ellipse is on the Y-axis, so the major axis of this ellipse will be the Y-axis.

Therefore, the equation of the ellipse will be of the form:

x2b2+y2a2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( length of semi-minor axis and foci) with the given one, we get

2b=16
b=8
and c=6

Now, as we know the relation,

a2=b2+c2

a=b2+c2

a=82+62

a=64+36

a=100

a=10

Hence, The Equation of the ellipse will be :

x282+y2103=1

x264+y2100=1 .

Question:17 Find the equation for the ellipse that satisfies the given conditions: Foci (± 3, 0), a = 4

Answer:

Given, in an ellipse, V Foci (± 3, 0), a = 4

Here, foci of the ellipse is in the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

So, on comparing standard parameters( vertices and foci) with the given one, we get

a=4 and c=3

Now, as we know the relation,

a2=b2+c2

b2=a2c2

b=a2c2

b=4232

b=7

Hence, The Equation of the ellipse will be :

x242+y2(7)2=1

x216+y27=1 .

Question:18 : Find the equation for the ellipse that satisfies the given conditions: b = 3, c = 4, centre at the origin; foci on the x-axis.

Answer:

Given: In an ellipse, b = 3, c = 4, centre at the origin; foci on the x-axis.

Here, foci of the ellipse is in the X-axis, so the major axis of this ellipse will be the X-axis.

Therefore, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

Also Given,

b=3 and c=4

Now, as we know the relation,

a2=b2+c2

a2=32+42

a2=25

a=5

Hence, The Equation of the ellipse will be :

x252+y232=1

x225+y29=1 .

Question:19: Find the equation for the ellipse that satisfies the given conditions: Centre at (0,0), major axis on the y-axis and passes through the points (3, 2) and (1,6).

Answer:

Given, in an ellipse

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

Since the major axis of this ellipse is on the Y-axis, the equation of the ellipse will be of the form:

x2b2+y2a2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

Now, since the ellipse passes through points (3, 2)

32b2+22a2=1

9a2+4b2=a2b2

Since the ellipse also passes through points (1, 6).

12b2+62a2=1

a2+36b2=a2b2

On solving these two equations, we get,

a2=40 and b2=10

Thus, the equation of the ellipse will be

x210+y240=1

Question:20: Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4,3) and (6,2).

Answer:

Given, in an ellipse

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

Since the major axis of this ellipse is on the X-axis, the equation of the ellipse will be of the form:

x2a2+y2b2=1

Where a and b are the length of the semimajor axis and semiminor axis, respectively.

Now, since the ellipse passes through the point (4,3)

42a2+32b2=1

16b2+9a2=a2b2

Since the ellipse also passes through the point (6, 2).

62a2+22b2=1

4a2+36b2=a2b2

On solving these two equations, we get,

a2=52 and b2=13

Thus, the equation of the ellipse will be

x252+y213=1


NCERT Conic Sections Class 11 Solutions: Exercise: 11.4
Page Number: 202
Total Questions: 15

Question:1 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: x216y29=1

Answer:
Given a Hyperbola equation, x216y29=1

It can also be written as x242y232=1

Comparing this equation with the standard equation of the hyperbola, we get,

x2a2y2b2=1

We get,

a=4 and b=3

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=42+32

c=5

Here, as we can see from the equation, the axis of the hyperbola is the X-axis.
So, the
Coordinates of the foci are:

(c,0) and (c,0)=(5,0) and (5,0)

The Coordinates of vertices:

(a,0) and (a,0)=(4,0) and (4,0)

The Eccentricity:

e=ca=54

The Length of the latus rectum :

2b2a=2(3)24=184=92

Question:2: Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: y29x227=1

Answer:

Given a Hyperbola equation, y29x227=1

It can also be written as y232x2(27)2=1

Comparing this equation with the standard equation of the hyperbola:

y2a2x2b2=1

We get,

a=3 and b=27

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=32+(27)2

c=36

c=6

Here, as we can see from the equation, the axis of the hyperbola is the Y-axis.
So, the
Coordinates of the foci are:

(0,c) and (0,c)=(0,6) and (0,6)

The Coordinates of vertices:

(0,a) and (0,a)=(0,3) and (0,3)

The Eccentricity:

e=ca=63=2

The Length of the latus rectum :

2b2a=2(27)3=543=18

Question:3 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: 9y24x2=36

Answer:

Given a Hyperbola equation, 9y24x2=36

It can also be written as

9y2364x236=1

y222x232=1

Comparing this equation with the standard equation of the hyperbola:

y2a2x2b2=1

We get,

a=2 and b=3

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=22+32

c=13

Hence, the Coordinates of the foci:

(0,c) and (0,c)=(0,13) and (0,13)

The Coordinates of vertices:

(0,a) and (0,a)=(0,2) and (0,2)

The Eccentricity:

e=ca=132

The Length of the latus rectum :

2b2a=2(9)2=182=9

Question:4 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: 16x29y2=576

Answer:

Given a Hyperbola equation, 16x29y2=576

It can also be written as:

16x25769y2576=1

x236y264=1

x262y282=1

Comparing this equation with the standard equation of the hyperbola:

x2a2y2b2=1

We get,

a=6 and b=8

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=62+82

c=10

Therefore, the Coordinates of the foci are:

(c,0) and (c,0)=(10,0) and (10,0)

The Coordinates of vertices:

(a,0) and (a,0)=(6,0) and (6,0)

The Eccentricity:

e=ca=106=53

The Length of the latus rectum :

2b2a=2(8)26=1286=643

Question:5 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: 5y29x2=36

Answer:

Given a Hyperbola equation, 5y29x2=36

It can also be written as:

5y2369x236=1

y2365x24=1

y2(65)2x222=1

Comparing this equation with the standard equation of the hyperbola:

y2a2x2b2=1

We get,

a=65 and b=2

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=(65)2+22

c=565

c=2145

Here, as we can see from the equation, the axis of the hyperbola is the Y-axis.
So, the Coordinates of the foci are:

(0,c) and (0,c)=(0,2145) and (0,2145)

The Coordinates of vertices:

(0,a) and (0,a)=(0,65) and (0,65)

The Eccentricity:

e=ca=214565=143

The Length of the latus rectum :

2b2a=2(4)65=453

Question:6 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas: 49y216x2=784

Answer:

Given a Hyperbola equation, 49y216x2=784

It can also be written as:

49y278416x2784=1

y216x249=1

y242x272=1

Comparing this equation with the standard equation of the hyperbola:

y2a2x2b2=1

We get,

a=4 and b=7

Now, as we know the relation in a hyperbola,

c2=a2+b2

c=a2+b2

c=42+72

c=65

Therefore,

Coordinates of the foci:

(0,c) and (0,c)=(0,65) and (0,65)

The Coordinates of vertices:

(0,a) and (0,a)=(0,4) and (0,4)

The Eccentricity:

e=ca=654

The Length of the latus rectum :

2b2a=2(49)4=984=492

Question:7 Find the equations of the hyperbola satisfying the given conditions: Vertices (± 2, 0), foci (± 3, 0)

Answer:

Given, in a hyperbola: Vertices (± 2, 0), foci (± 3, 0)

Here, Vertices and foci are on the X-axis, so the standard equation of the Hyperbola will be ;

x2a2y2b2=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=2 and c=3

Now, as we know, the relation in a hyperbola

c2=a2+b2

b2=c2a2

b2=3222

b2=94=5

Hence,The Equation of the hyperbola is x24y25=1.

Question:8 Find the equations of the hyperbola satisfying the given conditions: Vertices (0, ± 5), foci (0, ± 8)

Answer:

Given, in a hyperbola: Vertices (0, ± 5), foci (0, ± 8)

Here, Vertices and foci are on the Y-axis, so the standard equation of the Hyperbola will be:

y2a2x2b2=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=5 and c=8

Now, as we know, the relation in a hyperbola

c2=a2+b2

b2=c2a2

b2=8252

b2=6425=39

Hence, The Equation of the hyperbola is: y225x239=1.

Question:9 Find the equations of the hyperbola satisfying the given conditions: Vertices (0, ± 3), foci (0, ± 5)

Answer:

Given, in a hyperbola Vertices (0, ± 3), foci (0, ± 5)

Here, Vertices and foci are on the Y-axis, so the standard equation of the Hyperbola will be:

y2a2x2b2=1

By comparing the standard parameter (Vertices and foci) with the given one, we get

a=3 and c=5

Now, as we know, the relation in a hyperbola

c2=a2+b2

b2=c2a2

b2=5232

b2=259=16

Hence, The Equation of the hyperbola is y29x216=1.

Question:10: Find the equations of the hyperbola satisfying the given conditions: Foci (± 5, 0), the transverse axis is of length 8.

Answer:

Given, in a hyperbola: Foci (± 5, 0), the transverse axis is of length 8.

Here, the foci are on the X-axis, so the standard equation of the Hyperbola will be:

x2a2y2b2=1

By comparing the standard parameter (transverse axis length and foci) with the given one, we get

2a=8a=4 and c=5

Now, as we know, the relation in a hyperbola

c2=a2+b2

b2=c2a2

b2=5242

b2=2516=9

Hence, The Equation of the hyperbola is x216y29=1.

Question:11: Find the equations of the hyperbola satisfying the given conditions: Foci (0, ±13), the conjugate axis is of length 24.

Answer:

Given, in a hyperbola: Foci (0, ±13), the conjugate axis is of length 24.

Here, the foci are on the Y-axis, so the standard equation of the Hyperbola will be:

y2a2x2b2=1

By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get

2b=24b=12 and c=13

Now, as we know, the relation in a hyperbola

c2=a2+b2

a2=c2b2

a2=132122

a2=169144=25

Hence, The Equation of the hyperbola is y225x2144=1.

Question:12: Find the equations of the hyperbola satisfying the given conditions: Foci (±35,0), the latus rectum is of length 8.

Answer:

Given, in a hyperbola: Foci (±35,0), the latus rectum is of length 8.

Here, foci are on the X-axis, so the standard equation of the Hyperbola will be ;

x2a2y2b2=1

By comparing the standard parameter (length of latus rectum and foci) with the given one, we get

c=35 and
2b2a=8
2b2=8a
b2=4a

Now, as we know, the relation in a hyperbola

c2=a2+b2

c2=a2+4a

a2+4a=(35)2

a2+4a=45

a2+9a5a45=0

(a+9)(a5)=0

a=9 or, a=5

Since a can never be negative,

a=5

a2=25

So, b2=4a=4(5)=20

Hence, The Equation of the hyperbola is x225y220=1.

Question:13 Find the equations of the hyperbola satisfying the given conditions: Foci (± 4, 0), the latus rectum is of length 12

Answer:

Given, in a hyperbola: Foci (± 4, 0), the latus rectum is of length 12

Here, foci are on the X-axis, so the standard equation of the Hyperbola will be:

x2a2y2b2=1

By comparing the standard parameter (length of latus rectum and foci) with the given one, we get

c=4 and

2b2a=12
2b2=12a
b2=6a

Now, as we know, the relation in a hyperbola

c2=a2+b2

c2=a2+6a

a2+6a=42

a2+6a=16

a2+8a2a16=0

(a+8)(a2)=0

a=8 or, a=2

Since a can never be negative,

a=2

a2=4

So, b2=6a=6(2)=12

Hence, The Equation of the hyperbola is x24y212=1.

Question:14: Find the equations of the hyperbola satisfying the given conditions: vertices (± 7,0), e=43

Answer:

Given, in a hyperbola: vertices (± 7,0), and e=43

Here, Vertices is on the X-axis, so the standard equation of the Hyperbola will be:

x2a2y2b2=1

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get,

a=7 and

e=ca
43=c7
c=283

Now, as we know, the relation in a hyperbola

c2=a2+b2

b2=c2a2

b2=(283)272

b2=(7849)49

b2=(7844419)=3439

Hence, The Equation of the hyperbola is x2499y2343=1.

Question:15 Find the equations of the hyperbola satisfying the given conditions: Foci (0,±10), passing through (2,3)

Answer:

Given, in a hyperbola: Foci (0,±10), passing through (2,3)

Since the foci of the hyperbola is on the Y-axis, the equation of the hyperbola will be of the form:

y2a2x2b2=1

By comparing the standard parameter (foci) with the given one, we get

c=10

Now, as we know, in a hyperbola

a2+b2=c2

a2+b2=10(1)

Now, as the hyperbola passes through the point (2,3),

32a222b2=1

9b24a2=a2b2(2)

Solving Equations (1) and (2), we get,

9(10a2)4a2=a2(10a2)

a423a2+90=0

(a2)218a25a2+90=0

(a218)(a25)=0

a2=18 or, a=5

Now, as we know in a hyperbola c is always greater than a,
we choose the value a2=5

So, b2=10a2=105=5

Hence The Equation of the hyperbola is y25x25=1.

NCERT Conic Sections Class 11 Solutions: Exercise: Miscellaneous Exercise
Page Number: 204
Total Questions: 8

Question:1: If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.

Answer:

Let the parabolic reflector open towards the right.

So, the equation of the parabolic reflector will be y2=4ax

Now, since this curve will pass through the point (5,10), if we assume origin at the optical centre,

So,

102=4a(5)

a=10020=5

Hence, the focus of the parabola is (a,0)=(5,0).

Alternative Method:

As we know, on any concave curve, focus, f=R2

Focus, f=R2=102=5

Hence, the focus is 5 cm right to the optical centre.

Question:2: An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

Answer:

Since the Axis of the parabola is vertical, let the equation of the parabola be x2=4ay

it can be seen that this curve will pass through the point (52, 10) if we assume the origin at the bottom end of the parabolic arch.

So,

(52)2=4a(10)

a=25160=532

Hence, the equation of the parabola is:

x2=4×532×y

x2=2032y

x2=58y

Now, when y = 2 the value of x will be:

x=(58×2)=54=52

Hence, the width of the arch at this height is:

2x=2×52=5

Question:3 The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and 100 m long, is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the middle

Answer:

Given,

The width of the parabolic cable = 100m

The length of the shorter supportive wire attached = 6m

The length of the longer supportive wire attached = 30m

Since the rope opens upwards, the equation will be of the form x2=4ay

Now, if we consider the origin at the centre of the rope, the equation of the curve will pass through points: (50, 30-6) = (50,24)

502=4a×24

a=62524

Hence, the equation of the parabola is

x2=4×62524×y

x2=6256×y

Now, at a point 18 m right from the centre of the rope, the x coordinate of that point will be 18.
So, by the equation, the y-coordinate will be:

y=x24a=1824×625243.11 m

Hence, the length of the supporting wire attached to the roadway from the middle is
(3.11 + 6)m = 9.11 m

Question:4: An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

Answer:

The equation of the semi-ellipse will be of the form x2a2+y2b2=1,y>0

Now, according to the question,

the length of major axis = 2a = 8
a=4

The length of the semi-major axis = 2
b=2

Hence, the equation will be,

x242+y222=1,y>0

x216+y24=1,y>0

Now, at point 1.5 cm from the end, the x coordinate is 41.5=2.5

So, the height at this point is

(2.5)216+y24=1
y=4(12.5216)

y1.56 m

Hence, the height of the required point is 1.56 m.

Question:5: A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

Answer:

Let θ be the angle that the rod makes with the ground.

Now, at a point 3 cm from the end,

cosθ=x9

At the point of touching the ground

sinθ=y3

Now, as we know the trigonometric identity,

cos2θ+sin2θ=1

(x9)2+(y3)2=1

x281+y29=1

Hence, the equation is x281+y29=1

Question:6 Find the area of the triangle formed by the lines joining the vertex of the parabola x2=12y to the ends of its latus rectum.

Answer:

Given the parabola, x2=12y

Comparing this equation with x2=4ay, we get, a=3

Now, as we know, the coordinates of the ends of the latus rectum are:

(2a,a) and (2a,a)

So, the coordinates of the latus rectum are,

(2a,a) and (2a,a)=(6,3) and (6,3)

Now the area of the triangle with coordinates (0,0),(6,3) and (-6,3)

Width of the triangle =2×6=12

Height of the triangle = 3

So, the area of the triangle
= 12× Base × height
= 12×12×3
= 18

Hence, the required area is 18 square units.

Question:7 A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Answer:

As we know, if a point moves in a plane in such a way that its distance from two points remains constant, then the path is an ellipse.

Now, according to the question,

The distance between the point from where the sum of the distance from a point is constant = 10
2a=10
a=5

Now, the distance between the foci = 8
2c=8
c=4

Now, as we know the relation,

c2=a2b2

b2=a2c2

b=a2c2=5242=2516=9=3

Hence, the equation of the ellipse is:

x2a2+y2b2=1

x252+y232=1

x225+y29=1

Hence, the path of the man will be:

x225+y29=1

Question:8: An equilateral triangle is inscribed in the parabola y2=4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Answer:

Given an equilateral triangle inscribed in a parabola with the equation y2=4ax

The one coordinate of the triangle is A(0,0).

Now, let the other two coordinates of the triangle are

B(x,4ax) and C(x,4ax)

Now, since the triangle is equilateral,

BC=AB=CA

24ax=(x0)2+(4ax0)2

x2=12ax

x=12a

The coordinates of the points of the equilateral triangle are,

(0,0),(12,4a×12a),(12,4a×12a)
=(0,0),(12,43a) and (12,43a)

So, the side of the triangle is:

24ax=2×43a=83a

Here, students can check the solved examples before moving on to the exercises. Exercises are also prepared based on difficulty levels varying from easy to moderate level. These links are separate links to the exercises. Students can check each exercise separately and try to solve the next one on their own before checking the solutions.


Importance of solving NCERT Questions for Class 11 Chapter 10 Conic Sections

Class 11 maths chapter 10 question answers are essential for many things. But before solving them, strengthen your fundamentals of conic sections. Some important aspects of solving these problems are listed below.

  • A healthy number of questions come from this chapter in the final exams. So, by mastering this chapter, students can achieve higher grades.
  • This chapter is also the backbone of some other mathematical and science topics like coordinate geometry, calculus, and physics applications.
  • Questions from this chapter appear in important exams like JEE Mains, VITEE, and BITSAT. So, studying them beforehand helps to reduce the pressure of facing questions from this chapter during these exams.
  • The diagrams of conic sections help visualize the various parts of these shapes and their symmetry. This will be helpful for further studies like engineering, architecture, and data science.

All in all, mastering this chapter is an absolute necessity for students in class 11.

NCERT Solutions For Class 11: Subject Wise

Students can use the following links to check the solutions of the other subjects to achieve good marks in the exam.

NCERT Books and NCERT Syllabus

After studying the NCERT books for fundamental knowledge, students can solve problems from other books. Also, they can check the latest NCERT syllabus to stay updated using the following links.

Frequently Asked Questions (FAQs)

1. What are the important topics covered in NCERT Class 11 Maths Chapter 10?

Conic sections chapter in class 11 Covers the following topics:

  • Introduction
  • Sections of Cone
  • Circle
  • Parabola
  • Ellipse
  • Hyperbola

There are also many important sub-topics like latus rectum, standard equations of the conic sections, foci, major/minor axis, directrix, and eccentricity.

2. How to Solve Conic Section Problems from NCERT Class 11 Maths?

Before solving the conic sections problems, familiarise yourself with all the basics, general equations, key properties, and important formulae of this chapter. Then follow the following steps.

  • First, identify the type of standard equations given. Whether it is a circle, parabola, ellipse or hyperbola.
  • Use the necessary formulas to find properties like directrix, coordinates of foci, latus rectum, and eccentricity.
  • Analyze all the solved examples first before moving to the exercises.
  • Start with easier problems first and gradually increase the difficulty level of questions.
  • Draw the conic section to visualise its focus, directrix and axis, if necessary.
3. Where to Find Free NCERT Solutions for Class 11 Maths Chapter 10?

You can find the free NCERT solutions on NCERT's official websites.

You can also check these detailed free solutions prepared by experienced careers360 experts explaining every detail and providing extra information about the exercises. Check the following links.

4. What are the real-life applications of conic sections?

Here are some real-life uses of different conic sections.

CircleIt can be used in clocks, wheels, planetary motions, CDs & DVDs, etc.
ParabolaIt can be used in Satellite dishes, radio telescopes, Car headlights & flashlights, Projectile motions, Suspension bridges, etc.
EllipseIt can be used in planetary orbits, MRI, Whispering galleries, Car & train tracks, etc.
HyperbolaIt can be used in navigation, GPS systems, Nuclear radiation detection, optical lenses, cooling towers, boomerangs, telescopes, etc.
5. What is the difference between ellipse, parabola, and hyperbola in conic sections?
FeaturesEllipseParabolaHyperbola
DefinitionA set of points where the sum of the distances from two fixed foci is constant.A set of points equidistant from a fixed point called the focus and a fixed line called the directrix.A set of points where the absolute difference of distances from two fixed foci is constant.
ShapeOval shapedU-shapedTwo separate open curves(Look like two infinite bows)
Standard Equationx2a2+y2b2=1y2=4axx2a2y2b2=1
Foci(±c,0), where c2=a2b2(a,0)(±c,0), where c2=a2+b2
Eccentricity (e)0<e<1e=1e>1

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0.16\; J

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1.00\; J

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2.45×10−3 kg

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 6.45×10−3 kg

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200 \, \, J - 500 \, \, J

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0.02

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