Contents

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Conic Sections |

Exercise |
Ex 11.1, Ex 11.2, Ex 11.3, Ex 11.4 |

Number of Questions Solved |
62 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections

### Chapter 11 Conic Sections Exercise 11.1

**In each of the following Exercises 1 to 5, find the equation of the circle with**

**Question 1.**

centre (0, 2) and radius 2

**Solution:**

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

The equation of circle is,

(x-h)^{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

**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

**Question 3.**

centre and radius

**Solution:**

here h = , k = and r =

The equation of circle is,

**Question 4.**

centre (1, 1) and radius

**Solution:**

Here h = l, k=l and r =

The equation of circle is,

(x – h)^{2} + (y – k)^{2} = r^{2}

∴ (x – 1)^{2} + (y – 1)^{2} = ^{2}

⇒ x^{2} + 1 – 2x + y^{2} +1 – 2y = 2

⇒ x^{2} + y^{2} – 2x – 2y = 0

**Question 5.**

centre (-a, -b) and radius .

**Solution:**

Here h=-a, k = -b and r =

The equation of circle is, (x – h)^{2} + (y – k)^{2} = r^{2}

∴ (x + a)^{2} + (y + b)^{2} =

⇒ 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.**

**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 co-ordinates of the centre are (-5, 3) and radius is 6.

**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}=

Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we

have h = 2,k = 4 and r = .

Thus co-ordinates of the centre are (2, 4) and radius is .

**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} =

Comparing it with (x – h)^{2} + (y – k)^{2} = r^{2}, we have h = 4, k = -5 and r =

Thus co-ordinates of the centre are (4, -5) and radius is .

**Question 9.**

2x^{2} + 2y^{2} – x = 0

**Solution:**

The given equation of circle is,

2x^{2} + 2y^{2} – x = 0

**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.

**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 – 3y-11 = 0.

∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)

Solving (iv) and (v), we get

h = and k =

Putting these values of h and k in (ii), we get

⇒ ⇒

Thus required equation of circle is

⇒

⇒

⇒ 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.

**Question 12.**

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

**Solution:**

Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).

∴ Radius of circle

**Question 13.**

Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.

**Solution:**

Let the circle makes intercepts a with x-axis and b with y-axis.

∴ OA = a and OB = b

So the co-ordinates 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).

**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 co-ordinates of centre are (2, 2)

∴ radius of circle

**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)

= 4.3 < 5.

Thus the point (-2.5, 3.5) lies inside the circle.

### Chapter 11 Conic Sections Exercise 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.**

**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.

**Question 2.**

x^{2} = 6y

**Solution:**

The given equation of parabola is x^{2} = 6y which is of the form x^{2} = 4ay.

**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.

**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.

**Question 5.**

y^{2}= 10x

**Solution:**

The given equation of parabola is y^{2} = 10x, which is of the form y^{2} = 4ax.

**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:**

**Question 7.**

Focus (6, 0); directrix x = -6

**Solution:**

We are given that the focus (6, 0) lies on the x-axis, therefore x-axis 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.

**Question 8.**

Focus (0, -3); directri xy=3

**Solution:**

We are given that the focus (0, -3) lies on the y-axis, therefore y-axis 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.

**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 x-axis

∴ 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.

**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 x-axis

∴ 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.

**Question 11.**

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

**Solution:**

Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.

∴ The equation of the parabola is of the form y^{2} = 4ax

Since the parabola passes through point (2, 3)

**Question 12.**

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

**Solution:**

Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.

∴ The equation of the parabola is of the form x^{2} = 4ay

Since the parabola passes through point (5, 2)

### Chapter 11 Conic Sections Exercise 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.**

**Question 1.**

**Solution:**

Given equation of ellipse of

Clearly, 36 > 16

The equation of ellipse in standard form is

**Question 2.**

**Solution:**

Given equation of ellipse is

Clearly, 25 > 4

The equation of ellipse in standard form is

**Question 3.**

**Solution:**

Given equation of ellipse is

Clearly, 16 > 9

The equation of ellipse in standard form is

**Question 4.**

**Solution:**

Given equation of ellipse is

Clearly, 100 > 25

The equation of ellipse in standard form is

**Question 5.**

**Solution:**

Given equation of ellipse is

Clearly, 49 > 36

The equation of ellipse in standard form is

**Question 6.**

**Solution:**

Given equation of ellipse is

Clearly, 400 > 100

The equation of ellipse in standard form is

**Question 7.**

36x^{2} + 4y^{2} = 144

**Solution:**

Given equation of ellipse is 36x^{2} + 4y^{2} = 144

**Question 8.**

16x^{2} + y^{2} = 16

**Solution:**

Given equation of ellipse is16x^{2} + y^{2} = 16

**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:**

**Question 10.**

Vertices (±5, 0), foci (±4,0)

**Solution:**

Clearly, The foci (±4, o) lie on x-axis.

∴ The equation of ellipse is standard form is

**Question 11.**

Vertices (0, ±13), foci (0, ±5)

**Solution:**

Clearly, The foci (0, ±5) lie on y-axis.

∴ The equation of ellipse is standard form is

**Question 12.**

Vertices (±6, 0), foci (±4,0)

**Solution:**

Clearly, The foci (±4, 0) lie on x-axis.

∴ The equation of ellipse is standard form is

**Question 13.**

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

**Solution:**

Since, ends of major axis (±3, 0) lie on x-axis.

∴ The equation of ellipse in standard form

**Question 14.**

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

**Solution:**

Since, ends of major axis (0, ) lie on i-axis.

∴ The equation of ellipse in standard form

**Question 15.**

Length of major axis 26, foci (±5, 0)

**Solution:**

Since the foci (±5, 0) lie on x-axis.

∴ The equation of ellipse in standard form

**Question 16.**

Length of major axis 16, foci (0, ±6)

**Solution:**

Since the foci (0, ±6) lie on y-axis.

∴ The equation of ellipse in standard form

**Question 17.**

Foci (±3, 0) a = 4

**Solution:**

since the foci (±3, 0) on x-axis.

∴ The equation of ellipse in standard form

**Question 18.**

b = 3, c = 4, centre at the origin; foci on the x axis.

**Solution:**

Since the foci lie on x-axis.

∴ The equation of ellipse in standard form is

**Question 19.**

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

**Solution:**

Since the major axis is along y-axis.

∴ The equation of ellipse in standard form

**Question 20.**

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

**Solution:**

Since the major axis is along the x-axis.

∴ The equation of ellipse in standard form is

### Chapter 11 Conic Sections Exercise 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.**

**Question 1.**

**Solution:**

Given equation of hyperbola is

**Question 2.**

**Solution:**

Given equation of hyperbola is

**Question 3.**

9y^{2} – 4x^{2} = 36

**Solution:**

Given equation of hyperbola is9y^{2} – 4x^{2} = 36

**Question 4.**

16x^{2} – 9y^{2} = 576

**Solution:**

Given equation of hyperbola is16x^{2} – 9y^{2} = 576

**Question 5.**

5y^{2} – 9x^{2} = 36

**Solution:**

Given equation of hyperbola is

5y^{2} – 9x^{2} = 36

**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.**

**Question 7.**

Vertices (±2,0), foci (±3,0)

**Solution:**

Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form

**Question 8.**

Vertices (0, ±5), foci (0, ±8)

**Solution:**

Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form

**Question 9.**

Vertices (0, ±3), foci (0, ±5)

**Solution:**

Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

**Question 10.**

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

**Solution:**

Here foci are (±5, 0) which lie on x-axis. So the equation of the hyperbola in standard

**Question 11.**

Foci (0, ±13), the conjugate axis is of length 24.

**Solution:**

Here foci are (0, ±13) which lie on y-axis.

So the equation of hyperbola in standard

**Question 12.**

Foci (,0) , the latus rectum is of length 8.

**Solution:**

Here foci are (, 0) which lie on x-axis.

So the equation of the hyperbola in standard

**Question 13.**

Foci (±4, 0), the latus rectum is of length 12.

**Solution:**

Here foci are (±4, 0) which lie on x-axis.

So the equation of the hyperbola in standard

**Question 14.**

Vertices (+7, 0), e =

**Solution:**

Here vertices are (±7, 0) which lie on x-axis.

So, the equation of hyperbola in standard

**Question 15.**

Foci (0, ), passing through (2, 3).

**Solution:**

Here foci are (0, ) which lie on y-axis.

So the equation of hyperbola in standard form

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