NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 11 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Conic Sections |

Exercise |
Ex 11.2 |

Number of Questions Solved |
12 |

Category |
NCERT Solutions |

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

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