10 . CONIC SECTIONS

CBSE Class 11 Maths Chapter 10. Conic Sections

Chapter 10. CONIC SECTIONS

Class 11 Maths Chapter 10 Conic Sections Exercise 10.1 Solutions :

In each of the following Exercises 1 to 5, find the equation of the circle with
1. centre (0,2) and radius 2

2. centre (–2,3) and radius 4

3. Centre and radius

4. centre (1,1) and radius 2
5. centre and radius .

Solution : 1. centre (0,2) and radius 2

Here,

The equation of the required circle is

Therefore, the equation of the circle is .

2. centre (–2,3) and radius 4

Here,

The equation of the required circle is

Therefore, the equation of the circle is .

3. Centre and radius

Here,

The equation of the required circle is

Therefore, the equation of the circle is .

4. centre (1,1) and radius .

Here,

The equation of the required circle is

Therefore, the equation of the circle is .
5. centre and radius .

Here,

The equation of the required circle is

Therefore, the equation of the circle is .

In each of the following Exercises 6 to 9, find the centre and radius of the circles.
6.

7.
8.

Solution : 6. Given , the equation of the circle is

Comparing the general form of circle is, then and

Therefore, the centre is (– 5 , 3) and radius is 6 .

7. Given , the equation of the circle is

Comparing the general form of circle is , then and

Therefore, the centre is ( 2 , 4) and radius is .
8. Given , the equation of the circle is

Comparing the general form of circle is , then and

Therefore, the centre is ( 4 , – 5) and radius is .

9. Given , the equation of the circle is

Comparing the general form of circle is , then and

Therefore, the centre is and radius is .

10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line .

Solution : Let the equation of the circle be .

Since the circle passes through (4, 1) and (6,5), we have

and

From (i) and (ii) , we get

Also since the centre lies on the line , we have

Putting in (iii) , we get

We have ,
[from (i)]

The equation of the circle is

Hence, the equation of the required circle is

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

Solution : Let the equation of the circle be .

Since the circle passes through (2,3) and (– 1 ,1), we have

and

From (i) and (ii) , we get

Also since the centre lies on the line , we have

Putting in (iv) , we get

We have ,

The equation of the circle is

Hence, the equation of the required circle is

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

Solution : Let the equation of the circle be

Given, x-axis , i.e., centre and

Since the point (2,3) passing through (i) , we get

If , the

If , then

Therefore, the equation of the circle is or .

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

Solution : Let the equation of the circle is

Since , the circle passing through (0,0) , (a,0) and (0,b) we have

Again ,

and

Putting the value of and in (i) , we get

Therefore, the equation of the circle is

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

Solution : Let the equation of the circle is

Here , and

We have ,

Since the centre , we get

15. Does the point (–2.5, 3.5) lie inside, outside or on the circle ?

Solution : Given , the equation of the is

Here,

LHS :

Therefore, the point (–2.5, 3.5) lie inside the circle .

Class 11 Maths Chapter 10 Conic Sections Exercise 10.2 Solutions :

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.
1. 2. 3. 4. 5. 6.

Solution : 1.

The given equation involves , so the axis of symmetry is along the x-axis.

We have,

Here,

Focus , axis = X-axis

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

The given equation involves , so the axis of symmetry is along the y-axis.

We have ,

Here,

Focus , axis = Y -axis

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

The given equation involves , so the axis of symmetry is along the x-axis.

We have,

Here,

Focus , axis = X-axis (negative direction)

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

The given equation involves , so the axis of symmetry is along the y-axis.

We have ,

Here,

Focus , axis = Y -axis (negative direction)

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

The given equation involves , so the axis of symmetry is along the x-axis.

We have,

Here,

Focus , axis = X-axis

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

The given equation involves , so the axis of symmetry is along the y-axis.

We have ,

Here,

Focus , axis = Y -axis (negative direction)

Directrix,

The equation of latusrectum ,

Length of the of latusrectum

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

7. Focus (6,0); directrix = – 6

8. Focus (0,–3); directrix = 3
9. Vertex (0,0); focus (3,0)

10. Vertex (0,0); focus (–2,0)
11. Vertex (0,0) passing through (2,3) and axis is along -axis.
12. Vertex (0,0), passing through (5,2) and symmetric with respect to -axis.

7. Focus (6,0); directrix = – 6

Solution : Since, the focus (6,0) lies on the x-axis, the x-axis itself is the axis of the parabola.

The directrix, and the focus .

Hence the equation of the parabola is of the form .

Here,

Therefore, the equation is

8. Focus (0,–3); directrix = 3

Solution : Since, the focus (0, – 3) lies on the y-axis, the y-axis itself is the axis of the parabola.

The directrix, and the focus .

Hence the equation of the parabola is of the form .

Here,

Thetefore, the equation is
9. Vertex (0,0); focus (3,0)

Solution : Since, the focus (3,0) lies on the x-axis, the x-axis itself is the axis of the parabola.

The vertex = (0 , 0) and the focus .

Hence the equation of the parabola is of the form .

Here,

Therefore, the equation is

10. Vertex (0,0); focus (–2,0)
Solution : Since, the focus (– 2 ,0) lies on the x-axis, the x-axis itself is the axis of the parabola.

The vertex = (0 , 0) and the focus .

Hence, the equation of the parabola is of the form .

Here,

Therefore , the equation is

11. Vertex (0,0) passing through (2,3) and axis is along -axis.

Solution : Since, the parabola is symmetric about x-axis and the vertex (0,0) ,then

The equation is of the form or

But the parabola passes through (2,3) which lies in the first quadrant .

Thus the equation is of the form

Here,

Putting in (i) , we get

Therefore, the equation is

12. Vertex (0,0), passing through (5,2) and symmetric with respect to -axis.

Solution : Since, the parabola is symmetric about y-axis and the vertex (0,0) ,then

the equation is of the form or

But the parabola passes through (5 , 2) which lies in the first quadrant.

Thus the equation is of the form

Here,

Putting in(i) , we get

Therefore, the equation is

Class 11 Maths Chapter 10 Conic Sections Exercise 10.3 Solutions :

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.

1. 2. 3. 4. 5. 6. 7. 8. 9.

Solution : 1. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 2. We have ,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum
Solution : 3. We have ,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 4. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 5. We have ,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 6. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 7. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 8. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

Solution : 9. We have,

Here,

Since ,

Vertices

Eccentricity ,

Foci

The length of major axis

The length of minor axis

The length of the latus rectum

In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:
10. Vertices (± 5, 0), foci (± 4, 0)

Solution : Since the vertices are on x-axis, the equation will be of the form , where a is the semi-major axis.
Given , Vertices

And foci

We have,

Hence the equation of the ellipse is

11. Vertices (0, ± 13), foci (0, ± 5)

Solution : Since the vertices are on y-axis, the equation will be of the form

Given , Vertices

foci

We have,

So, the equation of the ellipse is

12. Vertices (± 6, 0), foci (± 4, 0)

Solution : Since the vertices (± 6, 0) and foci are on x-axis, the equation will be of the form
Given , Vertices

And foci

We have,

Hence the equation of the ellipse is

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

Solution : Since the major axis (± 3, 0) is along the X-axis. and the minor axis (0, ± 2) is along the Y-axis, the equation will be of the form

Given, the major axis

and the minor axis

Hence the equation of the ellipse is

14. Ends of major axis (0, ± ), ends of minor axis (± 1, 0)

Solution : Since the major axis (0 ,± ) is along the Y-axis and the minor axis ( ± 1 , 0) is along the X-axis, the equation will be of the form

Given, the major axis

and the minor axis

Hence the equation of the ellipse is

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

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

So, equation of the ellipse is of the form

Given, Length of major axis

And foci

We have,

Hence the equation of the ellipse is

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

Solution : Since the foci are on y-axis, the length of minor axis is along the y-axis.

So, equation of the ellipse is of the form

Given, The length of minor axis

And foci

We have,

Hence the equation of the ellipse is

17. Foci (± 3, 0), a = 4

Solution : Since the foci are on X-axis.

So, the equation of the ellipse is of the form

Foci

We have,

Hence the equation of the ellipse is

18. , centre at the origin; foci on the -axis.

Solution : Since, foci lies on the -axis.

So, the equation of the ellipse is of the form

Given,

And

We have,

Hence the equation of the ellipse is

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

Solution : Since the major axis on the -axis .

The standard form of the ellipse is

Since the points (3, 2) and (1, 6) lie on the ellipse, we have

And

Putting in (i) , we have

Hence the required equation is

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

Solution : Since the major axis on the -axis .

The standard form of the ellipse is

Since the points (4 , 3) and (6 , 2) lie on the ellipse, we have

And

[From (i)]

Putting in (i) , we get

Hence the required equation is

Class 11 Maths Chapter 10 Conic Sections Exercise 10.4 Solutions :

In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

Solution : 1. Given the equation of hyperbola is

Here , and

Since, the transverse axis is along X-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity,

Foci

The length of the latus rectum

Solution : 2.

Given the equation of hyperbola is

Here , and

Since, the transverse axis is along Y-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity,

Foci

The length of the latus rectum

Solution : 3.

Given the equation of hyperbola is

Here , and

Since, the transverse axis is along Y-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity,

Foci

The length of the latus rectum

Solution : 4.

Given the equation of hyperbola is

Here , and

Since, the transverse axis is along X-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity,

Foci

The length of the latus rectum

Solution : 5.

Given the equation of hyperbola is

Here , and

Since, the transverse axis is along Y-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity

Foci

The length of the latus rectum

Solution : 6.

Given the equation of hyperbola is

Here, and

Since, the transverse axis is along Y-axis ( because the coefficient of of the hyperbola is positive.)

Vertices

Eccentricity

Foci

The length of the latus rectum

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

7. Vertices (± 2, 0), foci (± 3, 0)

Solution : Since the foci is on X-axis, the equation of the hyperbola is of the form

Given, Vertices

And foci

Hence , the equation of the hyperbola is
8. Vertices (0, ± 5), foci (0, ± 8)

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

Given, Vertices

And foci

We have,

Hence , the equation of the hyperbola is
9. Vertices (0, ± 3), foci (0, ± 5)
Solution : Since the foci is on Y-axis, the equation of the hyperbola is of the form

Given, Vertices

And foci

We have,

Hence , the equation of the hyperbola is

10. Foci (± 5, 0), the transverse axis is of length 8.

Solution : Since the foci (± 5, 0) is lies on X-axis, the equation of the hyperbola is of the form

Foci

The length of transverse axis

We have,

Hence , the equation of the hyperbola is
11. Foci (0, ±13), the conjugate axis is of length 24 .

Solution : Since the foci (0 , ± 13) is lies on Y-axis, the equation of the hyperbola is of the form

Foci

The length of conjugate axis

We have,

Hence, the equation of the hyperbola is
12. Foci ( ), the latus rectum is of length 8.

Solution : Since the foci () is lies on X-axis, the equation of the hyperbola is of the form

Foci

The length of transverse axis

We have,

[From (i)]

⇒a=-9 (negative value not excepted )

Putting in (i) , we get

Hence , the equation of the hyperbola is
13. Foci (± 4, 0), the latus rectum is of length 12

Solution : Since the foci (± 5, 0) is lies on X-axis, the equation of the hyperbola is of the form

Foci

The length of the latus rectum

We have,

[ From (i) ]

(Negative value)

Putting in (i) , we have

Hence , the equation of the hyperbola is
14. Vertices ,

Solution : Since the vertices is lies on X-axis, the equation of the hyperbola is of the form

Vertices

Eccentricity,

We have,

Hence , the equation of the hyperbola is

i.e.,

15. Foci , passing through (2,3)

Solution : Since the vertices is lies on Y-axis, the equation of the hyperbola is of the form

Foci

Equation (i) , passing through the point (2 , 3) , then

We have,

(negative value not excepted)

Putting in (i) , we get

Hence , the equation of the hyperbola is

I.e.,

Class 11 Maths Chapter 10 Conic Sections Miscellaneous Exercise Solutions :

1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
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?
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.
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.
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 -axis.
6. Find the area of the triangle formed by the lines joining the vertex of the parabola to the ends of its latus rectum.
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.
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.