CBSE NCERT Solutions for Class 11 Mathematics Chapter 11€¦ · Practice more on Conic Sections...

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CBSE NCERT Solutions for Class 11 Mathematics Chapter 11

Back of Chapter Questions

Exercise: 11.1

1. Find the equation of the circle with centre and radius

Hint: )

Solution:

Solution Step1: As we all know that, the equation of a circle is given as

Given :- centre and radius

Now, the equation of the given circle can be written as:-


Hint: )

Solution:

Solution Step1: The equation of a circle is given as

Given: centre and radius

Therefore, the equation of the circle is


Hint:

Solution:

Solution Step1: The equation of a circle with centre and radius is given as

It is given that centre and radius

Now, the equation of the circle can be written as




Hint:

Solution:



Now, the equation of the circle can be written as


Hint:

Solution:



Therefore, the equation of the circle is



6. Find the centre and radius of the circle

Hint:

Solution:

Solution Step1: The equation of the given circle is

which is of the form where and

Thus, the centre of the given circle is while its radius is


Hint:

Solution:


Which is of the form where and



Hint:

Solution:







Hint:

Solution:


which is of the form where and Thus, the centre of the

given circle is while its radius is

10. Find the equation of the circle passing through the points and and whose centre is on the

line

Hint:

Solution:

Solution Step1: Let the equation of the required circle be

Since the circle passes through points and

…(i)

…(ii)

Since the centre of the circle lies on line

…(iii)

From equations (i) and (ii), we get

…(iv)



On solving equations (iii) and (iv), we get and

On substituting the values of and in equation (i), we get

Thus, the equation of the required circle is

11. Find the equation of the circle passing through the points and and whose centre is on

the line

Hint:

Solution:


Since the circle passes through points and

…(i)

…(ii)

Since the centre of the circle lies on line

…(iii)

From equations (i) and (ii), we get

…(iv)

On solving equations (iii) and (iv), we get and

On substituting the values of and in equation (i), we obtain




12. Find the equation of the circle with radius whose centre lies on -axis and passes through the point

Hint:

Solution:


Since the radius of the circle is and its centre lies on the -axis, and

Now, the equation of the circle becomes

It is given that the circle passes through point

If then

If then

When the equation of the circle becomes

When the equation of the circle becomes

13. Find the equation of the circle passing through and making intercepts and on the coordinate axes.

Hint:



Solution:


Since the centre of the circle passes through

The equation of the circle now becomes

It is given that the circle makes intercepts and on the coordinate axes. This means that the

circlepasses through points and Therefore,

…(i)

…(ii)

From equation (i), we obtain

or

However, hence,

From equation (ii), we obtain

or

However, hence,


14. Find the equation of a circle with centre and passes through the point

Hint:

Solution:

Solution Step1: The centre of the circle is given as

Since the circle passes through point the radius of the circle is the distance between the

points and

Thus, the equation of the circle is



15. Does the point lie inside, outside or on the circle

Hint:

Solution:



Centre and radius

Now, the distance between point and centre

Since the distance between point and centre of the circle is less than the radius of

the circle, point is lies inside the circle.

Exercise: 11.2

1. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the

latus rectum for

Hint:

Solution:

Solution Step1: The given equation is

Here, the coefficient of is positive. Hence, the parabola opens towards the right.

On comparing this equation with we obtain

Coordinates of the focus

Since the given equation involves the axis of the parabola is the -axis.

Equation of directrix, i.e., i.e.,

Length of latus rectum




latus rectum for

Hint:

Solution:


Here, the coefficient of is positive. Hence, the parabola opens upwards.




Equation of directrix, i.e.,


3. Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of

thelatus rectum for

Hint:

Solution:


Here, the coefficient of is negative. Hence, the parabola opens towards the left.







latus rectum for

Hint:

Solution:




Here, the coefficient of is negative. Hence, the parabola opens downwards.







latus rectum for

Hint:

Solution:


Here, the coefficient of is positive. Hence, the parabola opens towards the right. On comparing this

equation with we obtain






latus rectum for

Hint:

Solution:


Here, the coefficient of is negative. Hence, the parabola opens downwards.








7. Find the equation of the parabola that satisfies the following conditions: Focus directrix

Hint:

Solution:

Solution Step1: Focus directrix,

Since the focus lies on the -axis, the -axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form or

It is also seen that the directrix, is to the left of the -axis, while the focus is to the right

of the -axis.

Hence, the parabola is of the form

Here,

Thus, the equation of the parabola is

8. Find the equation of the parabola that satisfies the following conditions: Focus directrix

Hint: equation of the parabola

Solution:

Solution Step1: Focus directrix

Since the focus lies on the -axis, the -axis is the axis of the parabola.

Therefore, the equation of the parabola is either of the form or

It is also seen that the directrix, is above the -axis, while the focus is below the -

axis. Hence, the parabola is of the form

Here,


9. Find the equation of the parabola that satisfies the following conditions: Vertex focus Hint: Equation of the parabola



Solution:

Step1: Vertex focus

Since the vertex of the parabola is and the focus lies on the positive -axis, -axis is the axis of

the parabola, while the equation of the parabola is of the form

Since the focus is

Thus, the equation of the parabola is i.e.,

10. Find the equation of the parabola that satisfies the following conditions: Vertex focus Hint: equation of the parabola

Solution:

Solution Step1: Vertex focus

Since the vertex of the parabola is and the focus lies on the negative -axis, -axis is the axis

of the parabola, while the equation of the parabola is of the form

Since the focus is

Thus, the equation of the parabola is i.e.,

11. Find the equation of the parabola that satisfies the following conditions: Vertex passing

through and axis is along -axis.

Hint:

Solution:

Solution Step1:

Since the vertex is and the axis of the parabola is the -axis, the equation of the parabola is

either of the form or

The parabola passes through point which lies in the first quadrant.

Therefore, the equation of the parabola is of the form while point must satisfy the

equation




12. Find the equation of the parabola that satisfies the following conditions: Vertex passing

through and symmetric with respect to -axis

Hint: the equation of the parabola

Solution:

Solution Step1: Since the vertex is and the parabola is symmetric about the -axis, the

equation of the parabola is either of the form or

The parabola passes through point which lies in the first quadrant.

Therefore, the equation of the parabola is of the form while point must satisfy the

equation


Exercise: 11.3

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

Hint:

Solution:


Here, the denominator of is greater than the denominator of

Therefore, the major axis is along the -axis, while the minor axis is along the -axis.

On comparing the given equation with

we obtain and

Therefore, the coordinates of the foci are and

The coordinates of the vertices are and

Length of major axis

Length of minor axis



Eccentricity,




Hint:

Solution:

Solution Step1: The given equation is or


Therefore, the major axis is along the -axis, while the minor axis is along the -axis. On comparing

the given equation with we obtain and

Therefore,

The coordinates of the foci are

and

The coordinates of the vertices are and



Eccentricity,


3. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis,the eccentricity


Hint:

Solution:






the given equation with we obtain and

Therefore, the coordinates of the foci are

The coordinates of the vertices are



Eccentricity,




Hint::

Solution:




the given equation with

we obtain and Therefore, the coordinates of the foci are


Length of major axis Length of minor axis

Eccentricity,






Hint:

Solution:





we obtain and





Eccentricity,


6. Find the coordinates of the foci, the vertices, the length of major axis, the minor axis,the eccentricity


Hint:

Solution:





we obtain and Therefore,







Eccentricity,

Length of latusrectum



Hint:

Solution:


It can be written as

or,

or, …(i)



On comparing equation (i) with

we obtain and

Therefore,





Eccentricity,






Hint:

Solution:



or,

or, …(i)



On comparing equation (i) with

we obtain and

Therefore,





Eccentricity,




Hint:

Solution:



or,



or, …(i)



thegiven equation with

we obtain and

Therefore,





Eccentricity,


10. Find the equation for the ellipse that satisfies the given conditions: Vertices foci

Hint:

Solution:

Solution Step1: Vertices foci

Here, the vertices are on the -axis.

Therefore, the equation of the ellipse will be of the form where is the semi-major axis.

Accordingly, and

It is known that

Thus, the equation of the ellipse is or




Hint:

Solution:




Accordingly, and

It is known that



Hint:

Solution:




Accordingly,

It is known that




13. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis ends

of minor axis

Hint:

Solution:

Solution Step1: Ends of major axis ends of minor axis Here, the major axis is along

the -axis.


Accordingly, and

Thus, the equation of the ellipse is i.e.,

14. Find the equation for the ellipse that satisfies the given conditions: Ends of major axis ends

of minor axis

Hint: .

Solution:

Solution Step1: Ends of major axis ends of minor axis

Here, the major axis is along the -axis. Therefore, the equation of the ellipse will be of the

form where is the semi-major axis.

Accordingly, and


15. Find the equation for the ellipse that satisfies the given conditions: Length of major axis foci

Hint:



Solution:

Solution Step1: Length of major axis foci


Accordingly, and

It is known that


16. Find the equation for the ellipse that satisfies the given conditions: Length of minor axis foci

Hint:

Solution:

Solution Step1: Length of minor axis foci

Since the foci are on the -axis, the major axis is along the -axis.


Accordingly, and

It is known that


17. Find the equation for the ellipse that satisfies the given conditions: Foci

Hint:



Solution:

Solution Step1: Foci



Accordingly, and

It is known that

Thus, the equation of the ellipse is

18. Find the equation for the ellipse that satisfies the given conditions: centre at the origin;

foci on the axis.

Hint:

Solution:

Solution Step1: It is given that centre at the origin; foci on the -axis.



Accordingly,

It is known that


19. Find the equation for the ellipse that satisfies the given conditions: Centre at major axis on the

-axis and passes through the points and

Hint:



Solution:

Solution Step1: Since the centre is at and the major axis is on the -axis, the equation of the

ellipse will be of the form

…(i)

Where, is the semi-major axis

The ellipse passes through points and Hence,

…(ii)

…(iii)

On solving equations (ii) and (iii), we obtain and


20. Find the equation for the ellipse that satisfies the given conditions: Major axis on the -axis and

passesthrough the points and

Hint:

Solution:

Solution Step1: Since the major axis is on the -axis, the equation of the ellipse will be of the form

…(i)

Where, is the semi-major axis

The ellipse passes through points and Hence,

…(ii)

…(iii)

On solving equations (ii) and (iii), we obtain and




Exercise: 11.4

1. Find the coordinates of the foci and the vertices, the eccentricity and the length of the lotus rectum of

the hyperbola

Hint:

Solution:


On comparing this equation with the standard equation of hyperbola i.e., we obtain

and

We know that

Therefore,



Eccentricity,


2. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum

of the hyperbola

Hint:

Solution:


On comparing this equation with the standard equation of hyperbola i.e., we obtain

and

We know that

Therefore,





Eccentricity,


3. Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of

the hyperbola

Hint:

Solution:



Or,

Or, …(i)

On comparing equation (i) with the standard equation of hyperbola i.e., we obtain

and

We know that

Therefore,



Eccentricity,



the hyperbola

Hint:

Solution:





…(i)


and

We know that

Therefore,



Eccentricity,


5. Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum

of the hyperbola

Hint: :

Solution:


…(i)


and

We know that





Eccentricity,



the hyperbola

Hint:

Solution:



Or,

Or, …(i)


and

We know that

Therefore,



Eccentricity,


7. Find the equation of the hyperbola satisfying the give conditions: Vertices foci

Hint:



Solution:



Therefore, the equation of the hyperbola is of the form

Since the vertices are

Since the foci are

We know that

Thus, the equation of the hyperbola is


Hint:

Solution:





Since the foci are

We know that



Hint:

Solution:







Since the foci are

We know that


10. Find the equation of the hyperbola satisfying the given conditions: Foci the transverse axis

is of length

Hint:

Solution:

Solution Step1: Foci the transverse axis is of length

Here, the foci are on the -axis.


Since the foci are

Since the length of the transverse axis is

We know that


11. Find the equation of the hyperbola satisfying the given conditions: Foci the conjugate axis

is of length

Hint:

Solution:

Solution Step1: Foci the conjugate axis is of length





Since the foci are

Since the length of the conjugate axis is

We know that


12. Find the equation of the hyperbola satisfying the given conditions: Foci the latus rectum

is of length

Hint:

Solution:

Solution Step1: Foci the latus rectum is of length


Therefore, the equation of the hyperbola is of the form Since the foci are


We know that

Since is non-negative,


13. Find the equation of the hyperbola satisfying the given conditions: Foci the latus rectum is

of length



Hint:

Solution:

Solution Step1: Foci the latus rectum is of length



Since the foci are


We know that

Since is non-negative,


14. Find the equation of the hyperbola satisfying the given conditions: Vertices

Hint:

Solution:

Solution Step1: Vertices




It is given that



We know that


15. Find the equation of the hyperbola satisfying the given conditions: Foci passing through

Hint:

Solution:

Solution Step1: Foci passing through



Since the foci are

We know that

…(i)

Since the hyperbola passes through point

…(ii)

From equations (i) and (ii), we obtain

or

In hyperbola, i.e.,


35


Miscellaneous

1. If a parabolic reflector is in diameter and deep, find the focus. Hint: This can be diagrammatically represented as

Solution:

Solution Step1: The origin of the coordinate plane is taken at the vertex of the parabolic reflector in

such a way that the axis of the reflector is along the positive -axis. This can be diagrammatically

represented as

The equation of the parabola is of the form (as it is opening to the right).Since the parabola

passes through point

Therefore, the focus of the parabola is which is the mid-point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

2. An arch is in the form of a parabola with its axis vertical. The arch is high and wide at

thebase. How wide is it from the vertex of the parabola?

Hint:

Solution:

Solution Step1: The origin of the coordinate plane is taken at the vertex of the arch in such a way that

its vertical axis isalong the positive -axis.

This can be diagrammatically represented as


36


The equation of the parabola is of the form (as it is opening upwards).

It can be clearly seen that the parabola passes through point

Therefore, the arch is in the form of a parabola whose equation is

When

Hence, when the arch is from the vertex of the parabola, its width is approximately

3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway

which is horizontal and long is supported by vertical wires attached to the cable, the

longest wire being and the shortest being Find the length of a supporting wire attached

to the roadway from the middle.

Hint: diagrammatically represented as

Solution:

Solution Step1: The vertex is at the lowest point of the cable. The origin of the coordinate plane is

taken as the vertex of the parabola, while its vertical axis is taken along the positive -axis.This can

be diagrammatically represented as


37


Here, and are the longest and the shortest wires, respectively, attached to the cable.

is thesupporting wire attached to the roadway, from the middle.

Here, and

The equation of the parabola is of the form (as it is opening upwards).

The coordinates of point are

Since is a point on the parabola,

Equation of the parabola

The -coordinate of point is

Hence, at

(approx)

Thus, the length of the supporting wire attached to the roadway from the middle is

approximately

4. An arch is in the form of a semi-ellipse. It is wide and high at the centre. Find the height

of the arch at a point from one end.

Hint:

Solution:

Solution Step1: Since the height and width of the arc from the centre is and respectively, it

is clear that the length of the major axis is while the length of the semi-minor axis is

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken

along the -axis. Hence, the semi-ellipse can be diagrammatically represented as


38


The equation of the semi-ellipse will be of the form where is the semi-major axis

Accordingly,

Therefore, the equation of the semi-ellipse is …(i)

Let be a point on the major axis such that

Draw

The -coordinate of point is

On substituting the value of with in equation (i), we obtain

(approx.)

Thus, the height of the arch at a point from one end is approximately

5. A rod of length moves with its ends always touching the coordinate axes. Determine the

equationof the locus of a point on the rod, which is from the end in contact with the -

axis.

Hint:

Solution:

Solution Step1: Let be the rod making an angle with and be the point on it such that

Then,


39


From draw and

In

In

Since,

Or,

Thus, the equation of the locus of point on the rod is

6. Find the area of the triangle formed by the lines joining the vertex of the parabola to trends of its lotus rectum.

Hint: comparing this equation

Solution:

Solution Step1: The given parabola is


The coordinates of foci are

Let be the latus rectum of the given parabola.

The given parabola can be roughly drawn as


40


At

The coordinates of are while the coordinates of are

Therefore, the vertices of are and

Area of

Thus, the required area of the triangle is

7. A man running a racecourse notes that the sum of the distances from the two flag posts form him

isalways and the distance between the flag posts is Find the equation of the posts

traced by the man.

Hint:

Solution:

Solution Step1: Let and be the positions of the two flag posts and be the position of the

man.

Accordingly,

We know that if a point moves in a plane in such a way that the sum of its distances from two fixed

pointsis constant, then the path is an ellipse and this constant value is equal to the length of the major

axis of the ellipse.


41


Therefore, the path described by the man is an ellipse where the length of the major axis is

while points and are the foci.

Taking the origin of the coordinate plane as the centre of the ellipse, while taking the major axis

along the -axis, the ellipse can be diagrammatically represented as

The equation of the ellipse will be of the form where is the semi-major axis

Accordingly,

Distance between the foci

On using the relation

Thus, the equation of the path traced by the man is

8. An equilateral triangle is inscribed in the parabola where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Hint:

Solution:

Solution Step1: Let be the equilateral triangle inscribed in parabola

Let intersect the -axis at point


42


Let

From the equation of the given parabola, we have

The respective coordinates of points and are and

Since is an equilateral triangle,

Thus, the side of the equilateral triangle inscribed in parabola is


43


CBSE NCERT Solutions for Class 11 Mathematics Chapter 11€¦ · Practice more on Conic Sections...

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