NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections: As the name suggests a conic section is a curve obtained from the intersection of the surface of a cone with a plane. There are three types of conic section hyperbola, the parabola, and the ellipse. The circle is a special case of the ellipse which has been discussed in this chapter. In the NCERT solutions for class 11 maths chapter 11 conic sections, you will see problems related to abovementioned curves like circles, parabolas, hyperbolas and ellipses. The names hyperbola and parabola are given by Apollonius. This is an important chapter for you in CBSE class 11 final exam. This chapter is more important for competitive exams like JEE Mains, JEEAdvanced, VITEEE, BITSAT etc because every year many questions are asked from this topic. There is a total of 62 questions are given in 4 exercises of NCERT textbook. All these questions are explained in the CBSE NCERT solutions for class 11 maths chapter 11 conic sections. These curves have applications in fields like the design of antennas and telescopes, planetary motion, reflectors in automobile headlights, etc. There are 8 questions in a miscellaneous exercise. In the solutions of NCERT for class 11 maths chapter 11 conic sections, you will get solutions to miscellaneous exercise too. Check all NCERT solutions from class 6 to 12 which are explained in detail. There are four exercises and a miscellaneous exercise of this chapter are explained below.
Topics of NCERT grade 11 maths chapter11 Conic Sections
11.1 Introduction
11.2 Sections of a Cone
11.3 Circle
11.4 Parabola
11.5 Ellipse
11.6 Hyperbola
The complete solutions of NCERT class 11 mathematics chapter 11 is provided below:
NCERT solutions for class 11 maths chapter 11 conic sectionsExercise: 11.1
Question:1 Find the equation of the circle with
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
So Given Here
AND
So the equation of the circle is:
,
Question:2 Find the equation of the circle with
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
So Given Here
AND
So the equation of the circle is:
,
Question:3 Find the equation of the circle with
Answer:
As we know,
The equation of the circle with center ( h, k) and radius r is give by ;
So Given Here
AND
So the equation of circle is:
,
Question:4 Find the equation of the circle with
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
So Given Here
AND
So the equation of the circle is:
,
Question:5 Find the equation of the circle with
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
So Given Here
AND
So the equation of the circle is:
,
Question:6 Find the centre and radius of the circles.
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given here
Can also be written in the form
So, from comparing, we can see that
Hence the Radius of the circle is 6.
Question:7 Find the centre and radius of the circles.
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given here
Can also be written in the form
So, from comparing, we can see that
Hence the Radius of the circle is .
Question:8 Find the centre and radius of the circles.
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given here
Can also be written in the form
So, from comparing, we can see that
Hence the radius of the circle is .
Question:9 Find the centre and radius of the circles.
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given here
Can also be written in the form
So, from comparing, we can see that
Hence Center of the circle is the Radius of the circle is .
Question:10 Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line .
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given Here,
Condition 1: the circle passes through points (4,1) and (6,5)
Here,
Now, Condition 2:centre is on the line .
From condition 1 and condition 2
Now lets substitute this value of h and k in condition 1 to find out r
So now, the Final Equation of the circle is
Question:11 Find the equation of the circle passing through the points (2,3) and (–1,1) and hose centre is on the line .
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
Given Here,
Condition 1: the circle passes through points (2,3) and (–1,1)
Here,
Now, Condition 2: centre is on the line.
From condition 1 and condition 2
Now let's substitute this value of h and k in condition 1 to find out
So now, the Final Equation of the circle is
Answer:
As we know,
The equation of the circle with centre ( h, k) and radius r is given by ;
So let the circle be,
Since it's radius is 5 and its centre lies on xaxis,
And Since it passes through the point (2,3).
When ,The equation of the circle is :
When The equation of the circle is :
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,
Now since this circle passes through (0,0)
Now, this circle makes an intercept of a and b on the coordinate axes.it means circle passes through the point (a,0) and (0,b)
So,
Since
Similarly,
Since b is not equal to zero.
So Final equation of the Circle ;
Question:14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).
Answer:
Let the equation of circle be :
Now, since the centre of the circle is (2,2), our equation becomes
Now, Since this passes through the point (4,5)
Hence The Final equation of the circle becomes
Question:15 Does the point (–2.5, 3.5) lie inside, outside or on the circle ?
Answer:
Given, a circle
As we can see center of the circle is ( 0,0)
Now the distance between (0,0) and (–2.5, 3.5) is
Since distance between the given point and center of the circle is less than radius of the circle, the point lie inside the circle.
Solutions of NCERT for class 11 maths chapter 11 conic sectionsExercise: 11.2
Answer:
Given, a parabola with equation
This is parabola of the form which opens towards the right.
So,
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is XAxis.
The equation of the directrix
The length of the latus rectum:
.
Answer:
Given, a parabola with equation
This is parabola of the form which opens upward.
So,
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is YAxis.
The equation of the directrix
The length of the latus rectum:
.
Answer:
Given, a parabola with equation
This is parabola of the form which opens towards left.
So,
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is XAxis.
The equation of the directrix
The length of the latus rectum:
.
Answer:
Given, a parabola with equation
This is parabola of the form which opens downwards.
So,
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is YAxis.
The equation of the directrix
The length of the latus rectum:
.
Answer:
Given, a parabola with equation
This is parabola of the form which opens towards the right.
So,
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is XAxis.
The equation of the directrix
The length of the latus rectum:
.
Answer:
Given, a parabola with equation
This is parabola of the form which opens downwards.
So
By comparing the given parabola equation with the standard equation, we get,
Hence,
Coordinates of the focus :
Axis of the parabola:
It can be seen that the axis of this parabola is YAxis.
The equation of the directrix
The length of the latus rectum:
.
Question:7 Find the equation of the parabola that satisfies the given conditions:
Answer:
Given, in a parabola,
Focus : (6,0) And Directrix :
Here,
Focus is of the form (a, 0), which means it lies on the Xaxis. And Directrix is of the form which means it lies left to the YAxis.
These are the condition when the standard equation of a parabola is.
Hence the Equation of Parabola is
Here, it can be seen that:
Hence the Equation of the Parabola is:
.
Question:8 Find the equation of the parabola that satisfies the given conditions:
Answer:
Given,in a parabola,
Focus : Focus (0,–3); directrix
Here,
Focus is of the form (0,a), which means it lies on the Yaxis. And Directrix is of the form which means it lies above XAxis.
These are the conditions when the standard equation of a parabola is .
Hence the Equation of Parabola is
Here, it can be seen that:
Hence the Equation of the Parabola is:
.
Question:9 Find the equation of the parabola that satisfies the given conditions:
Answer:
Given,
Vertex (0,0) And focus (3,0)
As vertex of the parabola is (0,0) and focus lies in the positive Xaxis, The parabola will open towards the right, And the standard equation of such parabola is
Here it can be seen that
So, the equation of a parabola is
.
Question:10 Find the equation of the parabola that satisfies the given conditions:
Answer:
Given,
Vertex (0,0) And focus (2,0)
As vertex of the parabola is (0,0) and focus lies in the negative Xaxis, The parabola will open towards left, And the standard equation of such parabola is
Here it can be seen that
So, the equation of a parabola is
.
Question:11 Find the equation of the parabola that satisfies the given conditions:
Vertex (0,0) passing through (2,3) and axis is along x axis.
Answer:
Given
The Vertex of the parabola is (0,0).
The parabola is passing through (2,3) and axis is along the x axis, it will open towards right. and the standard equation of such parabola is
Now since it passes through (2,3)
So the Equation of Parabola is ;
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 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 Y=axis, it's axis will ve Yaxis. and since it passes through the point (5,2), it must go through the first quadrant.
So the standard equation of such parabola is
Now since this parabola is passing through (5,2)
Hence the equation of the parabola is
CBSE NCERT solutions for class 11 maths chapter 11 conic sectionsExercise: 11.3
Question:1 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.
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Xaxis and the minor axis is along Yaxis.
On comparing the given equation with the standard equation of an ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Yaxis and the minor axis is along Xaxis.
On comparing the given equation with the standard equation of such ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Xaxis and the minor axis is along Yaxis.
On comparing the given equation with the standard equation of an ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Yaxis and the minor axis is along Xaxis.
On comparing the given equation with the standard equation of such ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of ellipse
As we can see from the equation, the major axis is along Xaxis and the minor axis is along Yaxis.
On comparing the given equation with standard equation of ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Yaxis and the minor axis is along Xaxis.
On comparing the given equation with the standard equation of such ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Yaxis and the minor axis is along Xaxis.
On comparing the given equation with the standard equation of such ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Yaxis and the minor axis is along Xaxis.
On comparing the given equation with the standard equation of such ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
Answer:
Given
The equation of the ellipse
As we can see from the equation, the major axis is along Xaxis and the minor axis is along Yaxis.
On comparing the given equation with the standard equation of an ellipse, which is
We get
and .
So,
Hence,
Coordinates of the foci:
The vertices:
The length of the major axis:
The length of minor axis:
The eccentricity :
The length of the latus rectum:
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 in Xaxis so the major axis of this ellipse will be Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
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 in Yaxis so the major axis of this ellipse will be Yaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
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 in Xaxis so the major axis of this ellipse will be Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
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 Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Hence, The Equation of the ellipse will be :
Which is
.
Question:14 Find the equation for the ellipse that satisfies the given conditions:
Ends of major axis (0, ± ), ends of minor axis (± 1, 0)
Answer:
Given, In an ellipse,
Ends of the major axis (0, ± ), ends of minor axis (± 1, 0)
Here, the major axis of this ellipse will be Yaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Hence, The Equation of the ellipse will be :
Which is
.
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 in Xaxis so the major axis of this ellipse will be Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
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 Yaxis so the major axis of this ellipse will be Yaxis.
Therefore, the equation of the ellipse will be of the form:
Where and are the length of the semimajor axis and semiminor axis respectively.
So on comparing standard parameters( length of semiminor axis and foci) with the given one, we get
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
Question:17 Find the equation for the ellipse that satisfies the given conditions:
Answer:
Given, In an ellipse,
V Foci (± 3, 0), a = 4
Here foci of the ellipse are in Xaxis so the major axis of this ellipse will be Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and 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
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
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 are in Xaxis so the major axis of this ellipse will be Xaxis.
Therefore, the equation of the ellipse will be of the form:
Where and are the length of the semimajor axis and semiminor axis respectively.
Also Given,
and
Now, As we know the relation,
Hence, The Equation of the ellipse will be :
Which is
.
Question:19 Find the equation for the ellipse that satisfies the given conditions:
Centre at (0,0), major axis on the yaxis and passes through the points (3, 2) and (1,6).
Answer:
Given,in an ellipse
Centre at (0,0), major axis on the yaxis and passes through the points (3, 2) and (1,6).
Since, The major axis of this ellipse is on the Yaxis, the equation of the ellipse will be of the form:
Where and are the length of the semimajor axis and semiminor axis respectively.
Now since the ellipse passes through points,(3, 2)
since the ellipse also passes through points,(1, 6).
On solving these two equation we get
and
Thus, The equation of the ellipse will be
.
Question:20 Find the equation for the ellipse that satisfies the given conditions:
Major axis on the xaxis and passes through the points (4,3) and (6,2) .
Answer:
Given, in an ellipse
Major axis on the xaxis and passes through the points (4,3) and (6,2).
Since The major axis of this ellipse is on the Xaxis, the equation of the ellipse will be of the form:
Where and are the length of the semimajor axis and semiminor axis respectively.
Now since the ellipse passes through the point,(4,3)
since the ellipse also passes through the point (6,2).
On solving this two equation we get
and
Thus, The equation of the ellipse will be
NCERT solutions for class 11 maths chapter 11 conic sectionsExercise: 11.4
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Here as we can see from the equation that the axis of the hyperbola is X axis. So,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
Question:2 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Here as we can see from the equation that the axis of the hyperbola is Yaxis. So,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Hence,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Therefore,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Here as we can see from the equation that the axis of the hyperbola is Yaxis. So,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
Answer:
Given a Hyperbola equation,
Can also be written as
Comparing this equation with the standard equation of the hyperbola:
We get,
and
Now, As we know the relation in a hyperbola,
Therefore,
Coordinates of the foci:
The Coordinates of vertices:
The Eccentricity:
The Length of the latus rectum :
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 focii are on the Xaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (Vertices and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Hence,The Equation of the hyperbola is ;
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 focii are on the Yaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (Vertices and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Hence, The Equation of the hyperbola is ;
.
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 focii are on the Yaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (Vertices and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Hence, The Equation of the hyperbola is ;
.
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, focii are on the Xaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (transverse axis length and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Hence, The Equation of the hyperbola is ;
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, focii are on the Yaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (length of conjugate axis and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Hence, The Equation of the hyperbola is ;
.
Question:12 Find the equations of the hyperbola satisfying the given conditions.
Foci , the latus rectum is of length 8 .
Answer:
Given, in a hyperbola
Foci , the latus rectum is of length 8.
Here, focii are on the Xaxis so, the standard equation of the Hyperbola will be ;
By comparing standard parameter (length of latus rectum and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Since can never be negative,
Hence, The Equation of the hyperbola is ;
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, focii are on the Xaxis so, the standard equation of the Hyperbola will be ;
By comparing standard parameter (length of latus rectum and foci) with the given one, we get
and
Now, As we know the relation in a hyperbola
Since can never be negative,
Hence, The Equation of the hyperbola is ;
Question:14 Find the equations of the hyperbola satisfying the given conditions.
Answer:
Given, in a hyperbola
vertices (± 7,0), And
Here, Vertices is on the Xaxis so, the standard equation of the Hyperbola will be ;
By comparing the standard parameter (Vertices and eccentricity) with the given one, we get
and
From here,
Now, As we know the relation in a hyperbola
Hence, The Equation of the hyperbola is ;
Question:15 Find the equations of the hyperbola satisfying the given conditions.
Answer:
Given, in a hyperbola,
Foci , passing through (2,3)
Since foci of the hyperbola are in Yaxis, the equation of the hyperbola will be of the form ;
By comparing standard parameter (foci) with the given one, we get
Now As we know, in a hyperbola
Now As the hyperbola passes through the point (2,3)
Solving Equation (1) and (2)
Now, as we know that in a hyperbola is always greater than, we choose the value
Hence The Equation of the hyperbola is
NCERT solutions for class 11 maths chapter 11 conic sectionsMiscellaneous Exercise
Question:1 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
Answer:
Le the parabolic reflector opens towards the right.
So the equation of parabolic reflector will be,
Now, Since this curve will pass through the point (5,10) if we assume origin at the optical centre,
So
Hence, The focus of the parabola is,
.
Alternative Method,
As we know on any concave curve
Hence, Focus
.
Hence the focus is 5 cm right to the optical centre.
Answer:
Since the Axis of the parabola is vertical, Let the equation of the parabola be,
it can be seen that this curve will pass through the point (5/2, 10) if we assume origin at the bottom end of the parabolic arch.
So,
Hence, the equation of the parabola is
Now, when y = 2 the value of x will be
Hence the width of the arch at this height is
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 towards upwards, the equation will be of the form
Now if we consider origin at the centre of the rope, the equation of the curve will pass through points, (50,306)=(50,24)
Hence the equation of the parabola is
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 ycoordinate will be
Hence the length of the supporting wire attached to roadway from the middle is 3.11+6=9.11m.
Answer:
The equation of the semiellipse will be of the form
Now, According to the question,
the length of major axis = 2a = 8
The length of the semimajor axis =2
Hence the equation will be,
Now, at point 1.5 cm from the end, the x coordinate is 41.5 = 2.5
So, the height at this point is
Hence the height of the required point is 1.56 m.
Answer:
Let be the angle that rod makes with the ground,
Now, at a point 3 cm from the end,
At the point touching the ground
Now, As we know the trigonometric identity,
Hence the equation is,
Answer:
Given the parabola,
Comparing this equation with , we get
Now, As we know the coordinates of ends of latus rectum are:
So, the coordinates of latus rectum are,
Now the area of the triangle with coordinates (0,0),(6,3) and (6,3)
Widht of the triangle = 2*6=12
Height of the triangle = 3
So The area =
Hence the required area is 18 unit square.
Answer:
As we know that if a point moves in a plane in such a way that its distance from twopoint remain 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
Now, the distance between the foci=8
Now, As we know the relation,
Hence the equation of the ellipse is,
Hence the path of the man will be
Answer:
Given, an equilateral triangle inscribed in parabola with the equation.
The one coordinate of the triangle is A(0,0).
Now, let the other two coordinates of the triangle are
and
Now, Since the triangle is equilateral,
The coordinates of the points of the equilateral triangle are,
So, the side of the triangle is
NCERT solutions for class 11 mathematics
NCERT solutions for class 11 Subject wise
Important points of NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

Circle It is the set of all points in the plane that are equidistant(or the same distance) from a fixed point in the plane. The fixed point is the centre of the circle and the distance from the centre to a point on the circle is the radius of the circle. The equation of a circle with centre (h, k) and the radius r is

Parabola It is the set of all points in the plane that are equidistant(or the same distance) from a fixed line and a fixed point (not on the line) in the plane. The fixed point F is the focus of the parabola and the fixed line is called the directrix of the parabola. If the coordinates of focus(F) is (a, 0) a > 0 and directrix x = – a, then the equation of the parabola is

Ellipse It is formed by a point, which moves in a plane in such a manner that the sum of its distances from two fixed points in a plane is constant. The two fixed points in the plane are called the foci of the ellipse. If the foci of the ellipse are on the xaxis, then the equation of an ellipse is

Hyperbola It is the locus of a point in a plane, which moves in such a manner so that the difference of whose distances from two fixed points in the plane is constant. The two fixed points in the hyperbola are called the foci of the hyperbola. If the foci of the hyperbola are on the xaxis, then the equation of a hyperbola is
Tip You should remember standard equations and formulas for all the standard curves and try to solve problems from NCERT including miscellaneous exercise. If you are not able to do, you can take help from the solutions of NCERT for class 11 maths chapter 11 conic section.
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