NCERT Solutions  Class 11 Maths Chapter 11  Conic sections
CBSE Solutions  Maths Class 11
Check the below NCERT Solutions for Class 11 Maths Chapter 11 Conic sections Pdf free download. NCERT Solutions Class 11 Maths were prepared based on the latest exam pattern. We have Provided Conic sections Class 11 Maths NCERT Solutions to help students understand the concept very well.
NCERT  Class 11 Maths
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Maths 
Chapter:  11 
Chapters Name:  Conic sections 
Medium:  English 
Conic sections  Class 11 Maths  NCERT Books Solutions
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1
In each of the following Exercises 1 to 5, find the equation of the circle with
Ex 11.1 Class 11 Maths Question 1.
centre (0, 2) and radius 2
Solution.
Here h = 0,k = 2 and r = 2
The equation of circle is,
(xh)^{2} + (y k)^{2} = r^{2}
∴ (x – 0)^{2} + (y – 2)^{2} = (2)^{2}
⇒ x^{2} + y^{2} + 4 – 4y = 4
⇒ x^{2} + y^{2} – 4y = 0
Ex 11.1 Class 11 Maths Question 2.
centre (2,3) and radius 4
Solution.
Here h=2,k = 3 and r = 4
The equation of circle is,
(x – h)^{2} + (y – k)^{2} = r^{2}
∴(x + 2)^{2} + (y – 3)^{2} = (4)^{2}
⇒ x^{2} + 4 + 4x + y^{2} + 9 – 6y = 16
⇒ x^{2} + y^{2} + 4x – 6y – 3 = 0
Ex 11.1 Class 11 Maths Question 3.
centre \(\left( \frac { 1 }{ 2 } ,\quad \frac { 1 }{ 4 } \right) \) and radius \(\frac { 1 }{ 12 } \)
Solution.
here h = \(\frac { 1 }{ 2 } \), k = \(\frac { 1 }{ 4 } \) and r = \(\frac { 1 }{ 12 } \)
The equation of circle is,
Ex 11.1 Class 11 Maths Question 4.
centre (1, 1) and radius \(\sqrt { 2 } \)
Solution.
Here h = l, k=l and r = \(\sqrt { 2 } \)
The equation of circle is,
(x – h)^{2} + (y – k)^{2} = r^{2}
∴ (x – 1)^{2} + (y – 1)^{2} = \(\left( \sqrt { 2 } \right) \)^{2}
⇒ x^{2} + 1 – 2x + y^{2} +1 – 2y = 2
⇒ x^{2} + y^{2} – 2x – 2y = 0
Ex 11.1 Class 11 Maths Question 5.
centre (a, b) and radius \(\sqrt { { a }^{ 2 }{ b }^{ 2 } } \).
Solution.
Here h=a, k = b and r = \(\sqrt { { a }^{ 2 }{ b }^{ 2 } } \)
The equation of circle is, (x – h)^{2} + (y – k)^{2} = r^{2}
∴ (x + a)^{2} + (y + b)^{2} = \(\left( \sqrt { { a }^{ 2 }{ b }^{ 2 } } \right) \)
⇒ x^{2} + a^{2} + 2ax + y^{2} + b^{2} + 2by = a^{2} b^{2}
⇒ x^{2} + y^{2} + 2ax + 2 by + 2b^{2} = 0
In each of the following exercises 6 to 9, find the centre and radius of the circles.
Ex 11.1 Class 11 Maths Question 6.
(x + 5)^{2} + (y – 3)^{2} = 36
Solution.
The given equation of circle is,
(x + 5)^{2} + (y – 3)^{2} = 36
⇒ (x + 5)^{2} + (y – 3)^{2} = (6)^{2}
Comparing it with (x – h)2^{2} + (y – k)^{2} = r^{2}, we get
h = 5, k = 3 and r = 6.
Thus the coordinates of the centre are (5, 3) and radius is 6.
Ex 11.1 Class 11 Maths Question 7.
x^{2} + y^{2} – 4x – 8y – 45 = 0
Solution.
The given equation of circle is
x^{2} + y^{2} – 4x – 8y – 45 = 0
∴ (x^{2} – 4x) + (y^{2} – 8y) = 45
⇒ [x^{2} – 4x + (2)^{2}] + [y^{2} – 8y + (4)^{2}] = 45 + (2)^{2} + (4)^{2}
⇒ (x – 2)^{2} + (y – 4)^{2} = 45 + 4 + 16
⇒ (x – 2)^{2} + (y – 4)^{2} = 65
⇒ (x – 2)^{2} + (y – 4)^{2}= \(\left( \sqrt { 65 } \right) ^{ 2 }\)
Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we
have h = 2,k = 4 and r = \(\sqrt { 65 } \).
Thus coordinates of the centre are (2, 4) and radius is \(\sqrt { 65 } \).
Ex 11.1 Class 11 Maths Question 8.
x^{2} + y^{2} – 8x + 10y – 12 = 0
Solution.
The given equation of circle is,
x^{2} + y^{2} – 8x + 10y 12 = 0
∴ (x^{2} – 8x) + (y^{2} + 10y) = 12
⇒ [x^{2} – 8x + (4)^{2}] + [y^{2} + 10y + (5)^{2}] = 12 + (4)^{2} + (5)^{2}
⇒ (x – 4)^{2} + (y + 5)^{2} = 12 + 16 + 25
⇒ (x – 4)^{2} + (y + 5)^{2} = 53
⇒ (x – 4)^{2} + (y + 5)^{2} = \(\left( \sqrt { 53 } \right) ^{ 2 }\)
Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we have h = 4, k = 5 and r = \(\sqrt { 53 } \)
Thus coordinates of the centre are (4, 5) and radius is \(\sqrt { 53 } \).
Ex 11.1 Class 11 Maths Question 9.
2x^{2} + 2y^{2} – x = 0
Solution.
The given equation of circle is,
2x^{2} + 2y^{2} – x = 0
Ex 11.1 Class 11 Maths 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.
Solution.
The equation of the circle is,
(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)^{2} + (1 – k)^{2} = r^{2}
⇒ 16 + h^{2} – 8h + 1 + k^{2} – 2k = r^{2}
⇒ h^{2}+ k^{2} – 8h – 2k + 17 = r^{2} …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h^{2} + (5 – k)^{2} = r^{2}
⇒ 36 + h^{2} 12h + 25 + k^{2} – 10k = r^{2}
⇒ h^{2} + k^{2} – 12h – 10kk + 61 = r^{2} …. (iii)
From (ii) and (iii), we have h^{2} + k^{2} – 8h – 2k +17
= h^{2} + k^{2}– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)^{2} + (4)^{2} – 8 x 3 – 2 x 4 + 17 = r^{2}
∴ r^{2} = 10
Thus required equation of circle is
(x – 3)^{2} + (y – 4)^{2} = 10
⇒ x^{2} + 9 – 6x + y^{2} +16 – 8y = 10
⇒ x^{2} + y^{2} – 6x – 8y +15 = 0.
Ex 11.1 Class 11 Maths Question 11.
Find the equation of the circle passing through the points (2, 3) and (1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution.
The equation of the circle is,
(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)^{2} + (3 – k)^{2} = r^{2}
⇒ 4 + h^{2} – 4h + 9 + k^{2} – 6k = r^{2}
⇒ h^{2}+ k^{2} – 4h – 6k + 13 = r^{2} ….(ii)
Also, the circle passes through point (1, 1)
∴ (1 – h)^{2} + (1 – k)^{2} = r^{2}
⇒ 1 + h^{2} + 2h + 1 + k^{2} – 2k = r^{2}
⇒ h^{2} + k^{2} + 2h – 2k + 2 = r^{2} ….(iii)
From (ii) and (iii), we have
h^{2} + k^{2} – 4h – 6k + 13 = h^{2} + k^{2} + 2h – 2k + 2
⇒ 6h – 4k = 11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y11 = 0.
∴ h – 3k – 11 = 0 ⇒ h 3k = 11 …(v)
Solving (iv) and (v), we get
h = \(\frac { 7 }{ 2 } \) and k = \(\frac { 5 }{ 2 } \)
Putting these values of h and k in (ii), we get
\(\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { 5 }{ 2 } \right) ^{ 2 }\frac { 4\times 7 }{ 2 } 6\times \frac { 5 }{ 2 } +13={ r }^{ 2 }\)
⇒ \(\frac { 49 }{ 4 } +\frac { 25 }{ 4 } 14+15+13 \) ⇒ \({ r }^{ 2 }=\frac { 65 }{ 2 } \)
Thus required equation of circle is
⇒ \(\left( x\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 } \)
⇒ \({ x }^{ 2 }+\frac { 49 }{ 4 } 7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 } \)
⇒ 4x^{2} + 49 – 28x + 4y^{2} + 25 + 20y = 130
⇒ 4x^{2} + 4y^{2} – 28x + 20y – 56 = 0
⇒ 4(x^{2} + y^{2} – 7x + 5y 14) = 0
⇒ x^{2} + y^{2} – 7x + 5y 14 = 0.
Ex 11.1 Class 11 Maths Question 12.
Find the equation of the circle with radius 5 whose centre lies on xaxis and passes through the point (2, 3).
Solution.
Since the centre of the circle lies on xaxis, the coordinates of centre are (h, 0). Now the circle passes through the point (2, 3).
∴ Radius of circle
Ex 11.1 Class 11 Maths Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Solution.
Let the circle makes intercepts a with xaxis and b with yaxis.
∴ OA = a and OB = b
So the coordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).
Ex 11.1 Class 11 Maths Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution.
The equation of circle is
(x – h)^{2} + (y – k)^{2} = r^{2} ….(i)
Since the circle passes through point (4, 5) and coordinates of centre are (2, 2)
∴ radius of circle
Ex 11.1 Class 11 Maths Question 15.
Does the point (2.5, 3.5) lie inside, outside or on the circle x^{2} + y^{2} = 25?
Solution.
The equation of given circle is x^{2} + y^{2} = 25
⇒ (x – 0)^{2} + (y – 0)^{2} = (5)^{2}
Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we
get
h = 0,k = 0, and r = 5
Now, distance of the point (2.5, 3.5) from the centre (0, 0)
\(\sqrt { \left( 0+2.5 \right) ^{ 2 }+\left( 03.5 \right) ^{ 2 } } =\sqrt { 6.25+12.25 } \)
\(\sqrt { 18.5 } \) = 4.3 < 5.
Thus the point (2.5, 3.5) lies inside the circle.
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2
In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.
Ex 11.2 Class 11 Maths Question 1.
y^{2}= 12x
Solution.
The given equation of parabola is y^{2} = 12x which is of the form y^{2} = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.
Ex 11.2 Class 11 Maths Question 2.
x^{2} = 6y
Solution.
The given equation of parabola is x^{2} = 6y which is of the form x^{2} = 4ay.
Ex 11.2 Class 11 Maths Question 3.
y^{2} = – 8x
Solution.
The given equation of parabola is
y^{2} = 8x, which is of the form y^{2} = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.
Ex 11.2 Class 11 Maths Question 4.
x^{2} = 16y
Solution.
The given equation of parabola is
x^{2} = 16y, which is of the form x^{2} = 4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, 4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.
Check the Parabola Calculator to solve Parabola Equation.
Ex 11.2 Class 11 Maths Question 5.
y^{2}= 10x
Solution.
The given equation of parabola is y^{2} = 10x, which is of the form y^{2} = 4ax.
Ex 11.2 Class 11 Maths Question 6.
x^{2} = 9y
Solution.
The given equation of parabola is
x^{2} = 9y, which is of the form x^{2} = 4ay.
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Ex 11.2 Class 11 Maths Question 7.
Focus (6, 0); directrix x = 6
Solution.
We are given that the focus (6, 0) lies on the xaxis, therefore xaxis is the axis of parabola. Also, the directrix is x = 6 i.e. x = a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y^{2} = 4ax.
The required equation of parabola is
y^{2} = 4 x 6x ⇒ y^{2} = 24x.
Ex 11.2 Class 11 Maths Question 8.
Focus (0, 3); directri xy=3
Solution.
We are given that the focus (0, 3) lies on the yaxis, therefore yaxis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, 3) i.e. (0, a). The equation of parabola is of the form x^{2} = 4ay.
The required equation of parabola is
x^{2} = – 4 x 3y ⇒ x^{2} = 12y.
Ex 11.2 Class 11 Maths Question 9.
Vertex (0, 0); focus (3, 0)
Solution.
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along xaxis
∴ The equation of the parabola is of the form y^{2} = 4ax
The required equation of the parabola is
y^{2} = 4 x 3x ⇒ y^{2} = 12x.
Ex 11.2 Class 11 Maths Question 10.
Vertex (0, 0); focus (2, 0)
Solution.
Since the vertex of the parabola is at (0, 0) and focus is at (2, 0).
∴ y = 0 ⇒ The axis of parabola is along xaxis
∴ The equation of the parabola is of the form y^{2} = – 4ax
The required equation of the parabola is
y^{2} = – 4 x 2x ⇒ y^{2} = 8x.
Ex 11.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along xaxis.
Solution.
Since the vertex of the parabola is at (0, 0) and the axis is along xaxis.
∴ The equation of the parabola is of the form y^{2} = 4ax
Since the parabola passes through point (2, 3)
Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect toyaxis.
Solution.
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the yaxis.
∴ The equation of the parabola is of the form x^{2} = 4ay
Since the parabola passes through point (5, 2)
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3
In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
Ex 11.3 Class 11 Maths Question 1.
\(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1\)
Solution.
Given equation of ellipse of \(\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1\)
Clearly, 36 > 16
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 2.
\(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1\)
Solution.
Given equation of ellipse is \(\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1\)
Clearly, 25 > 4
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 3.
\(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Solution.
Given equation of ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Clearly, 16 > 9
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 4.
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1\)
Solution.
Given equation of ellipse is \(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1\)
Clearly, 100 > 25
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 5.
\(\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1\)
Solution.
Given equation of ellipse is \(\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1\)
Clearly, 49 > 36
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 6.
\(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Solution.
Given equation of ellipse is \(\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1\)
Clearly, 400 > 100
The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 7.
36x^{2} + 4y^{2} = 144
Solution.
Given equation of ellipse is 36x^{2} + 4y^{2} = 144
Ex 11.3 Class 11 Maths Question 8.
16x^{2} + y^{2} = 16
Solution.
Given equation of ellipse is16x^{2} + y^{2} = 16
Ex 11.3 Class 11 Maths Question 9.
4x^{2} + 9y^{2} = 36
Solution.
Given equation of ellipse is4x^{2} + 9y^{2} = 36
In each 0f the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:
Ex 11.3 Class 11 Maths Question 10.
Vertices (±5, 0), foci (±4,0)
Solution.
Clearly, The foci (±4, o) lie on xaxis.
∴ The equation of ellipse is standard form is
Ex 11.3 Class 11 Maths Question 11.
Vertices (0, ±13), foci (0, ±5)
Solution.
Clearly, The foci (0, ±5) lie on yaxis.
∴ The equation of ellipse is standard form is
Ex 11.3 Class 11 Maths Question 12.
Vertices (±6, 0), foci (±4,0)
Solution.
Clearly, The foci (±4, 0) lie on xaxis.
∴ The equation of ellipse is standard form is
Ex 11.3 Class 11 Maths Question 13.
Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Solution.
Since, ends of major axis (±3, 0) lie on xaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 14.
Ends of major axis (0, \(\pm \sqrt { 5 } \)), ends of minor axis (±1, 0)
Solution.
Since, ends of major axis (0, \(\pm \sqrt { 5 } \)) lie on iaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 15.
Length of major axis 26, foci (±5, 0)
Solution.
Since the foci (±5, 0) lie on xaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 16.
Length of major axis 16, foci (0, ±6)
Solution.
Since the foci (0, ±6) lie on yaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 17.
Foci (±3, 0) a = 4
Solution.
since the foci (±3, 0) on xaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 18.
b = 3, c = 4, centre at the origin; foci on the x axis.
Solution.
Since the foci lie on xaxis.
∴ The equation of ellipse in standard form is
Ex 11.3 Class 11 Maths Question 19.
Centre at (0, 0), major axis on the yaxis and passes through the points (3, 2) and (1, 6)
Solution.
Since the major axis is along yaxis.
∴ The equation of ellipse in standard form
Ex 11.3 Class 11 Maths Question 20.
Major axis on the xaxis and passes through the points (4, 3) and (6, 2).
Solution.
Since the major axis is along the xaxis.
∴ The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4
In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, eccentricity and the length of the latus rectum of the hyperbolas.
Ex 11.4 Class 11 Maths Question 1.
\(\frac { { x }^{ 2 } }{ 16 } \frac { { y }^{ 2 } }{ 9 } =1\)
Solution.
Given equation of hyperbola is
Ex 11.4 Class 11 Maths Question 2.
\(\frac { { y }^{ 2 } }{ 9 } \frac { x^{ 2 } }{ 27 } =1\)
Solution.
Given equation of hyperbola is
Ex 11.4 Class 11 Maths Question 3.
9y^{2} – 4x^{2} = 36
Solution.
Given equation of hyperbola is9y^{2} – 4x^{2} = 36
Ex 11.4 Class 11 Maths Question 4.
16x^{2} – 9y^{2} = 576
Solution.
Given equation of hyperbola is16x^{2} – 9y^{2} = 576
Ex 11.4 Class 11 Maths Question 5.
5y^{2} – 9x^{2} = 36
Solution.
Given equation of hyperbola is
5y^{2} – 9x^{2} = 36
Ex 11.4 Class 11 Maths Question 6.
49y^{2} – 16x^{2} = 784
Solution.
Given equation of hyperbola is49y^{2} – 16x^{2} = 784
In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.
Ex 11.4 Class 11 Maths Question 7.
Vertices (±2,0), foci (±3,0)
Solution.
Vertices are (±2, 0) which lie on xaxis. So the equation of hyperbola in standard form
Ex 11.4 Class 11 Maths Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution.
Vertices are (0, ±5) which lie on xaxis. So the equation of hyperbola in standard form
Ex 11.4 Class 11 Maths Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution.
Vertices are (0, ±3) which lie on xaxis. So the equation of hyperbola in standard form
Ex 11.4 Class 11 Maths Question 10.
Foci (±5, 0), the transverse axis is of length 8.
Solution.
Here foci are (±5, 0) which lie on xaxis. So the equation of the hyperbola in standard
Ex 11.4 Class 11 Maths Question 11.
Foci (0, ±13), the conjugate axis is of length 24.
Solution.
Here foci are (0, ±13) which lie on yaxis.
So the equation of hyperbola in standard
Ex 11.4 Class 11 Maths Question 12.
Foci (\(\pm 3\sqrt { 5 } \),0) , the latus rectum is of length 8.
Solution.
Here foci are (\(\pm 3\sqrt { 5 } \), 0) which lie on xaxis.
So the equation of the hyperbola in standard
Ex 11.4 Class 11 Maths Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution.
Here foci are (±4, 0) which lie on xaxis.
So the equation of the hyperbola in standard
Ex 11.4 Class 11 Maths Question 14.
Vertices (+7, 0), e = \(\frac { 4 }{ 3 } \)
Solution.
Here vertices are (±7, 0) which lie on xaxis.
So, the equation of hyperbola in standard
Ex 11.4 Class 11 Maths Question 15.
Foci (0, \(\pm \sqrt { 10 } \)), passing through (2, 3).
Solution.
Here foci are (0, \(\pm \sqrt { 10 } \)) which lie on yaxis.
So the equation of hyperbola in standard form
NCERT Class 11 Maths
Class 11 Maths Chapters  Maths Class 11 Chapter 11
Chapterwise NCERT Solutions for Class 11 Maths

NCERT Solutions For Class 11 Maths Chapter 1 Sets
NCERT Solutions For Class 11 Maths Chapter 2 Functions and relations
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric functions
NCERT Solutions For Class 11 Maths Chapter 4 Principle of mathematical induction
NCERT Solutions For Class 11 Maths Chapter 5 Quadratic equations and complex numbers
NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions For Class 11 Maths Chapter 7 Permutations and combinations
NCERT Solutions For Class 11 Maths Chapter 8 Binomial Theorem
NCERT Solutions For Class 11 Maths Chapter 9 Sequences and series
NCERT Solutions For Class 11 Maths Chapter 10 Straight lines
NCERT Solutions For Class 11 Maths Chapter 11 Conic sections
NCERT Solutions For Class 11 Maths Chapter 12 Introduction to threedimensional geometry
NCERT Solutions For Class 11 Maths Chapter 13 Limits and derivatives
NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning
NCERT Solutions For Class 11 Maths Chapter 15 Statistics
NCERT Solutions For Class 11 Maths Chapter 16 Probability
NCERT Solutions for Class 12 All Subjects  NCERT Solutions for Class 10 All Subjects 
NCERT Solutions for Class 11 All Subjects  NCERT Solutions for Class 9 All Subjects 
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