**NCERT solutions of class 11 maths exercise 11.2 -Conic sections**

All the **NCERT solutions of class 11 maths exercise 11.2 chapter 11-Conic sections** are created here are the solutions of unsolved questions of the **chapter 11-conic sections** which is one of the important chapter.The concept of the chapter **conic section of NCERT text book** is utilised in the industry and science and technology as an example planetry motion,in designing telescope,antennas,reflector in flashlights and automobile headlights.**Conic sections** are actually the surfaces generated by the intersection of a plane on a double napped right circular cone.

**Download pdf of NCERT solutions class 11 chapter 11-Conic Section**

**pdf-NCERT solutions class 11 chapter 11-Conic Section**

All the NCERT solutions are explained by an expert of maths,each NCERT solutions are explained here by a step by step method,so every students can understand it properly,it will help all the students of class 11 studens in boosting their preparation of fortcomming CBSE board exams.

**NCERT solutions of class 11 maths exercise 11.2 -Conic sections**

**Q1.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = 12x**

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Ans. The given equation is y² = 12x

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

On comparing the equation with y² = 4ax, we obtain

4a = 12

a = 12/4

a = 3

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

Since the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x = -a

Length of lactus rectum = 4a = 4 × 3 = 12

**Q2.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for x² = 6y**

Ans. The given equation is x² = 6y

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

On comparing the equation with x² = 4ay, we obtain

6y = 4ay

4a = 6

a = 6/4

a = 3/2

Coordinates of the focus = (0,a) = (0,3/2)

Since the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, y = -a

Length of lactus rectum = 4a = 6

**Q3.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = -8x**

Ans. The given equation is y² = -8x

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

On comparing the equation with y² = -4ax, we obtain

-4a = -8

a = 8/4

a = 2

Coordinates of the focus = (-a,o) = (-2,0)

Since the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x = a

Length of lactus rectum = 4a = 8

**Q4.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for x² = -16y**

Ans. The given equation is x² = -16y

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

On comparing the equation with x² = -4ay, we obtain

-4a = -16

a = 16/4

a = 4

Coordinates of the focus = (o,-a) = (0,-4)

Since the given equation involves x², the axis of the parabola is the y-axis.

Equation of directrix, y = a

Length of lactus rectum = 4a = 16

**Q5.Find the coordinates of the focus, axis of parabola, the equation of directrix and the length of the lactus rectum for y² = 10x**

Ans. The given equation is y² = 10x

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

On comparing the equation with y² = 4ax, we obtain

4a = 10

a = 10/4

a = 5/2

Coordinates of the focus = (a,o) = (5/2,0)

Since the given equation involves y², the axis of the parabola is the x-axis.

Equation of directrix, x = -a⇒ x = -5/2

Length of lactus rectum = 4a = 4×5/2 =10

**NCERT solutions of class 11 maths exercise 11.2 -Conic sections**

**Q6.Find the coordinates of the focus ,axis of parabola ,the equation of directrix and length of latus rectum for x² = -9y.**

Ans. The given equation of the parabola is x² = -9y

Comparing the given equation with the standard equation of parabola

x² = 4ay

4a = -9

a = -9/4

Coordinates of the focus = (a,0) =(-9/4, 0)

Since the given equation(x² =-9y) is symmetric about y-axis so the axis of parabola is y-axis

The equation of directrix= x = -a⇒x = -(-9/4) ⇒x =9/4

And length of latus rectum =4a = 4×9/4 =9

In each of the section of the exercises 7 to 12, find the equation of the parabola that satisfies the given conditions.

**Q7. Focus (6,0), directrix x = -6**

Ans. Focus of the parabola is : (a,0) and directrix x = -a

We are given here focus : (6,0)=(a,0)⇒ a = 6

directrix is x =-6 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens at left side of y-axis

So,the equation of parabola is y² = 4ax

Putting a = 6,we get

y² = 4×6x⇒ y² = 24x

y² = 24x

**Q8.Focus (0,-3): direcrix, y = 3**

Ans. Focus of the parabola is : (0,a) and directrix y= -a

We are given here focus : (0,a)=(0,-3)⇒ a = -3 and directrix y =3

directrix is y =3 indicates that the parabola is symmetric about y-axis and positive sign shows that it opens downwards of x -axis

So,the equation of parabola is x² = 4ay

Putting a = -3,we get

x² = 4×-3y⇒ x² = -12y

x² = -12y

**Q9.Vertex (0,0): focus (3,0)**

Ans. Focus of the parabola is : (a,0) and the coordinates of the vertex are (0,0)

We are given here focus : (a,0)=(3,0)⇒ a = 3 and directrix will be x= -a ⇒x = -3

directrix is x = -3 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis

So,the equation of parabola is y² = 4ax

Putting a = 3,we get

y² = 4×3x⇒ y² = 12x

y² = 12x

**Q10.Vertex (0,0) : focus (-2,0)**

Ans. Focus of the parabola is : (a,0) and the coordinates of the vertex are (0,0)

We are given here focus : (a,0)=(-2,0)⇒ a = -2 and directrix will be x= -a ⇒x = -(-2) = 2⇒ x = 2

directrix is x = 2 indicates that the parabola is symmetric about x-axis and positive sign shows that it opens left of y -axis

Equation of parabola is y² = 4ax

Putting a = -2,we get

y² = 4×-2x⇒ y² = -8x

y² = -8x

**Q11. Vertex (0,0) passing through (2,3) and axis is along x-axis.**

Ans. Since axis of parabola is on x-axis therefore equation of parabola is

y² =4ax, it is given to us that parabola is passing through (2,3)

Putting x =2,y =3 in the equation of parabola

3² = 4×a×2⇒a =9/8

The focus of parabola is : (a,0)=(9/8,0)⇒ a = 9/8 and directrix will be x= -a ⇒x = -9/8

directrix is x = -9/8 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis

So,the equation of parabola is

y² =4ax, putting a =9/8 in the equation

y² = 4×(9/8)x

2y² =9x

**NCERT solutions of class 11 maths exercise 11.2 -Conic sections**

**Q12.Vertex (0,0) passing through (5,2) and symmetric with respect to y-axis.**

Ans. Since axis of parabola is symmetric with respect to y-axis. therefore equation of parabola is

x² =4ay, it is given to us that parabola is passing through (5,2)

Putting x =5,y =2 in the equation of parabola

5² = 4×a×2 ⇒a =25/8

The focus of parabola is : (0,a)=(0,25/8)⇒ a = 25/8 and directrix will be y= -a⇒x = -25/8

directrix is x = -25/8 indicates that the parabola is symmetric about x-axis and negative sign shows that it opens right of y -axis

So,the equation of parabola is

x² =4ay, putting a =25/8 in the equation

x² = 4×(25/8)y

x² =(25/2)x ⇒2x² =25y

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Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

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Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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