# NCERT Solutions for Class 11 Maths Chapter 11

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections (Shanku Parichchhed) in PDF file format free download in English Medium updated for academic session 2020-21. Students can do directly 12th class after doing 10th though NIOS Admission 2020-2021.

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## NCERT Solutions for Class 11 Maths Chapter 11

Class: | 11 |

Subject: | Maths |

Chapter 11: | Conic Sections |

### 11th Maths Chapter 11 Solutions

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections all exercise and miscellaneous are given below to free download in PDF form updated for new academic session 2020-21. NCERT Solutions and Offline Apps 2020-21 are based on latest NCERT Books 2020-21 following new CBSE Syllabus 2020-2021.

#### Important Terms on Conic Sections

1. Circle, ellipse, parabola and hyperbola are curves which are obtained by intersection of a plane and cone in different positions

2. Circle: It is the set of all points in a plane that are equidistant from a fixed point in that plane.

3. Parabola: It is the set of all points in a plane which are equidistant from a fixed point (focus) and a fixed line in the plane. Fixed point does not lie on the line.

4. Latus Rectum: A chord through focus perpendicular to axis of parabola is called its latus rectum.

5. Ellipse: It is the set of points in a plane the sum of whose distances from two fixed points in the plane is a constant and is always greater than the distances between the fixed points.

6. Hyperbola: It is the set of all points in a plane, the differences of whose distance from two fixed points in the plane is a constant.

##### Questions from Exam Papers

1. Find the centre and radius of the circle x² + y² – 6x + 4y – 12 = 0. [Answer: (3, -2), 5]

2. Find the equation of hyperbola satisfying given conditions foci (5, 0) and transverse axis is of length 8. [Answer: x²/16 – y²/9 = 1.]

3. Find the coordinates of points on parabola y² = 8x whose focal distance is 4. [Answer: (2, 4), (2, -4)]

4. If one end of a diameter of the circle x² + y² – 4x – 6y + 11 = 0 is (3, 4), then find the coordinates of the other end of diameter. [Answer: (1, 2)]

5. Find equation of an ellipse having vertices (0, 5) and foci (0, 4). [Answer: x²/9 + y²/25 = 1]

6. Find the equation of a circle whose centre is at (4, –2) and 3x – 4y+ 5 = 0 is tangent to circle. [Answer: x² + y² – 8x + 4y – 5 = 0]

###### Questions for Practice

1. If the distance between the foci of a hyperbola is 16 and its eccentricity is 2, then obtain the equation of a hyperbola. [Answer: x² – y² = 32]

2. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity. [Answer: e = √3/2]

3. Find the length of major and minor axis of the following ellipse, 16x² + 25y² = 400. [Answer: 10, 8]

4. Find equation of circle concentric with circle 4x² + 4y² – 12x – 16y – 21 = 0 and of half its area. [Answer: 2x² + 2y² – 6x + 8y + 1 = 0]

5. Find the equation for the ellipse that satisfies the given condition Major axis on the x-axis and passes through the points (4, 3) and (6, 2). [Answer: x²/52 + y²/13 = 1]

### Important Questions on 11th Maths Chapter 11

(x – h)2 + (y – k)2 = r2

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)2 + (y – 2)2 = 22

⟹ x2 + y2 + 4 – 4 y = 4

⟹ x2 + y2 – 4y = 0

(x – h)^2 + (y – k)^2 = r^2

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2

⟹ x^2 + 4x + 4 + y^2 – 6y + 9 = 16

⟹ x^2 + y^2 + 4x – 6y – 3 = 0

(x + 5)^2 + (y – 3)^2 = 36

⇒ {x – (–5)}^2 + (y – 3)^2 = 62,

which is of the form (x – h)^2 + (y – k)^2 = r^2,

where h = – 5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

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

(4 – h)^2 + (1 – k)^2 = r^2 …… (1)

(6 – h)^2 + (5 – k)^2 = r^2 …… (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k = 16 …… (3)

From equations (1) and (2), we obtain

(4 – h)^2 + (1 – k)^2 = (6 – h)^2 + (5 – k)^2

⇒ 16 – 8h + h^2 + 1 – 2k + k^2 = 36 – 12h + h^2 + 25 – 10k + k^2

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44

⇒ h + 2k = 11 …… (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)^2 + (1 – 4)^2 = r^2

⇒ (1)^2 + (– 3)^2 = r^2

⇒ 1 + 9 = r^2

⇒ r^2 = 10

⇒ =√10

Thus, the equation of the required circle is

(x – 3)^2 + (y – 4)^2 = (√10)^2

x^2 – 6x + 9 + y^2 – 8y + 16 = 10

x^2 + y^2 – 6x – 8y + 15 = 0

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

On comparing this equation with y^2 = 4ax, we obtain

4a = 12

⇒ a = 3

∴ Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y^2, the axis of the parabola is the x-axis.

Equation of direcctrix,

x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12.

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

Therefore, the equation of the parabola is either of the form y2 = 4ax or

y^2 = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.

Hence, the parabola is of the form y^2 = 4ax.

Here, a = 6

Thus, the equation of the parabola is y^2 = 24x.