R S Aggarwal Solutions for Class 11 Maths Chapter 22 Parabola

R S Aggarwal Solutions for Chapter 22 help students to understand Parabola and terms related to it such as foci, vertices, axes, directrix, length of latus rectum and many more. To facilitate easy learning and help students understand the concepts of Parabola, free R S Aggarwal solutions are provided here.

Important points to remember:

1. The general form of a parabola: y2 = 4ax

Focus : F(a,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x + a = 0

Axis: y = 0

Length of latus rectum : 4a

2. The general form of a parabola: y2 = -4ax

Focus : F(-a,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x – a = 0

Axis: y = 0

Length of latus rectum : 4a

3. The general form of a parabola: x2 = 4ay

Focus : F(0,a)

Vertex : A(0,0) (at any point A)

Equation of the directrix : y + a = 0

Axis: x = 0

Length of latus rectum : 4a

4. The general form of a parabola: x2 = -4ay

Focus : F(0, -a)

Vertex : A(0,0) (at any point A)

Equation of the directrix : y – a = 0

Axis: x = 0

Length of latus rectum : 4a

Download PDF of R S Aggarwal Solutions for Class 11 Maths Chapter 22 Parabola

 

 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22
 rs aggarwal solution class 11 maths chapter 22

 

Access Answers to Maths R S Aggarwal Chapter 22 Parabola

Exercise 22 Page No: 741

Question 1: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola:

(i) y2 = 12x

(ii) y2 = 10x

(iii) 3y2 = 8x

Solution:

The general form of a parabola: y2 = 4ax ….(1)

Focus : F(a,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x + a = 0

Axis: y = 0

Length of latus rectum : 4a

(i) y2 = 12x

On comparing given equation with (1), we have

4a = 12 => a = 3

Now,

Focus : F(a,0) = F(3,0)

Vertex : A(0,0)

Equation of the directrix : x + a = 0

=> x + 3 = 0

or x = -3

Axis: y = 0

Length of latus rectum : 4a = 4 x 3 = 12 units

(ii) y2 = 10x

On comparing given equation with (1), we have

4a = 10 => a = 2.5

Now,

Focus : F(a,0) = F(2.5,0)

Vertex : A(0,0)

Equation of the directrix : x + a = 0

=> x + 2.5 = 0

or x = -2.5

Axis: y = 0

Length of latus rectum : 4a = 4 x (2.5) = 10 units

(iii) 3y2 = 8x

or y2 = 8/3 x

On comparing given equation with (1), we have

4a = 8/3 => a = 2/3

Now,

Focus : F(a,0) = F(2/3,0)

Vertex : A(0,0)

Equation of the directrix : x + a = 0

=> x + 2/3 = 0

or x = -2/3

Axis: y = 0

Length of latus rectum : 4a = 4 x 2/3 = 8/3 units

Question 2: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola:

(i) y2 = -8x

(ii) y2 = -6x

(iii) 5y2 = -16x

Solution:

The general form of a parabola: y2 = -4ax ….(1)

Focus : F(-a,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x – a = 0

Axis: y = 0

Length of latus rectum : 4a

(i) y2 = -8x

On comparing given equation with (1), we have

4a = 8 => a = 2

Now,

Focus : F(-2,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x – 2 = 0 or x = 2

Axis: y = 0

Length of latus rectum : 4a = 4 x 2 = 8 units

(ii) y2 = -6x

On comparing given equation with (1), we have

4a = 6 => a = 3/2

Now,

Focus : F(-3/2,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x – 3/2 = 0 or x = 3/2 or 2x – 3 = 0

Axis: y = 0

Length of latus rectum : 4a = 4 x 3/2 = 6 units

(iii) 5y2 = -16x

or y2 = -16/5 x

On comparing given equation with (1), we have

4a = 16/5 => a = 4/5

Now,

Focus : F(-4/5,0)

Vertex : A(0,0) (at any point A)

Equation of the directrix : x – 4/5 = 0 or 5x – 4 = 0

Axis: y = 0

Length of latus rectum : 4a = 4 x 4/5 = 16/5 units


Question 3: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola :

(i) x2 = 16y

(ii) x2 = 10y

(iii) 3x2 = 8y

Solution:

The general form of a parabola: x2 = 4ay ….(1)

Focus : F(0,a)

Vertex : A(0,0) (at any point A)

Equation of the directrix : y + a = 0

Axis: x = 0

Length of latus rectum : 4a

(i) x2 = 16y

On comparing given equation with (1), we have

4a = 16 => a = 4

Now,

Focus : F(0, 4)

Vertex : A(0, 0)

Equation of the directrix : y + 4 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 4 = 16 units

(ii) x2 = 10y

On comparing given equation with (1), we have

4a = 10 => a = 2.5

Now,

Focus : F(0, 2.5)

Vertex : A(0, 0)

Equation of the directrix : y + 2.5 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 2.5 = 10 units

(iii) 3x2 = 8y

or x2 = 8/3 y

On comparing given equation with (1), we have

4a = 8/3 => a = 2/3

Now,

Focus : F(0, 2/3)

Vertex : A(0, 0)

Equation of the directrix : y + 2/3 = 0 or 3y + 2 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 2/3 = 8/3 units

Question 4: Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola :

(i) x2 = -8y

(ii) x2 = -18y

(iii) 3x2 = -16y

Solution:

The general form of a parabola: x2 = -4ay ….(1)

Focus : F(0, -a)

Vertex : A(0,0) (at any point A)

Equation of the directrix : y – a = 0

Axis: x = 0

Length of latus rectum : 4a

(i) x2 = -8y

On comparing given equation with (1), we have

4a = 8 => a = 2

Now,

Focus : F(0, -2)

Vertex : A(0, 0)

Equation of the directrix : y – 2 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 2 = 8 units

(ii) x2 = -18y

On comparing given equation with (1), we have

4a = 18 => a = 9/2

Now,

Focus : F(0, -9/2)

Vertex : A(0, 0)

Equation of the directrix : y – 9/2 = 0 or 2y – 9 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 9/2 = 18 units

(iii) 3x2 = -16y

Or x2 = -16/3 y

On comparing given equation with (1), we have

4a = 16/3 => a = 4/3

Now,

Focus : F(0, -4/3)

Vertex : A(0, 0)

Equation of the directrix : y – 4/3 = 0 or 3y – 4 = 0

Axis: x = 0

Length of latus rectum : 4a = 4 x 4/3 = 16/3 units

Question 5: Find the equation of the parabola with vertex at the origin and focus at F(-2, 0).

Solution:

Give: Vertex : A (0,0) and focus, F(-2,0)

We know, For Vertex A(0,0) and Focus F(-a,0), equation of parabola is: y2 = – 4ax

Here, a = 2

Therefore, equation of parabola: y2 = – 8x

Question 6: Find the equation of the parabola with focus F(4, 0) and directrix x = -4.

Solution:

We are given with focus F(4, 0) and directrix x = -4 or x + 4 = 0

We know, For directrix with equation x + a = 0 and focus (a, 0), equation of the parabola is, y2 = 4ax

Here, a = 4

Therefore, equation of parabola: y2 = 16x

Question 7: Find the equation of the parabola with focus F(0, -3) and directrix y = 3.

Solution:

Given, focus F(0, -3) and directrix y = 3 or y – 3 = 0.

We know, For directrix with equation y – a = 0 and focus (0, -a), equation of the parabola is: x2 = – 4ay

Here, a = 3

Therefore, equation of parabola: x2 = – 12y

Question 8: Find the equation of the parabola with vertex at the origin and focus F(0, 5).

Solution:

We have to find equation of the parabola with origin and focus F(0, 5).

We know, For vertex A(0, 0) (origin at point A) and focus, F(0, a), equation of the parabola is: x2 = 4ay

Here, a = 5

Therefore, equation of parabola: x2 = 20y

Question 9: Find the equation of the parabola with vertex at the origin, passing through the point P(5, 2) and symmetric with respect to the y-axis.

Solution:

The equation of a parabola with vertex at the origin and symmetric about the y-axis: x2 = 4ay

As we are given, parabola is passing through the point P(5,2).

Putting x = 5 and y = 2 in x2 = 4ay

=> 25 = 4a(2) = 8a

=> a = 25/8

Therefore, equation of parabola:

x2 = 4(25/8 )y = 25/2 y

or 2x2 = 25y

Question 10: Find the equation of the parabola, which is symmetric about the y-axis and passes through the point P(2, -3).

Solution: The equation of a parabola with vertex at the origin and symmetric about the y-axis: x2 = 4ay

As we are given, parabola is passing through the point P(2, -3).

Putting x = 2 and y = -3 in x2 = 4ay

=> 4 = -12a

=> a = -1/3

Therefore, equation of parabola:

x2 = 4(-1/3)y = -4/3 y

or 3x2 = -4y


R S Aggarwal Solutions for Chapter 22 Parabola

R S Aggarwal Chapter 22 – Parabola, contains the answers to all the questions listed in this chapter. Let us have a look at some of the topics that are being covered here:

  • Parabola introduction
  • Intersection of nappes of a cone and a plane
  • Right-handed parabola
  • Length of latus rectum
  • Left-handed parabola
  • Upward parabola
  • Downward parabola

Download R S Aggarwal Class 11 solutions pdf and practice questions related to various concepts covered in this textbook.