- RD Chapter 1- Sets
- RD Chapter 2- Relations
- RD Chapter 3- Functions
- RD Chapter 4- Measurement of Angles
- RD Chapter 5- Trigonometric Functions
- RD Chapter 6- Graphs of Trigonometric Functions
- RD Chapter 7- Trigonometric Ratios of Compound Angles
- RD Chapter 8- Transformation Formulae
- RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles
- RD Chapter 10- Sine and Cosine Formulae and Their Applications
- RD Chapter 11- Trigonometric Equations
- RD Chapter 12- Mathematical Induction
- RD Chapter 13- Complex Numbers
- RD Chapter 14- Quadratic Equations
- RD Chapter 15- Linear Inequations
- RD Chapter 16- Permutations
- RD Chapter 17- Combinations
- RD Chapter 18- Binomial Theorem
- RD Chapter 19- Arithmetic Progressions
- RD Chapter 20- Geometric Progressions
- RD Chapter 21- Some Special Series
- RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates
- RD Chapter 23- The Straight Lines
- RD Chapter 24- The Circle
- RD Chapter 26- Ellipse
- RD Chapter 27- Hyperbola
- RD Chapter 28- Introduction to 3D coordinate geometry
- RD Chapter 29- Limits
- RD Chapter 30- Derivatives
- RD Chapter 31- Mathematical Reasoning
- RD Chapter 32- Statistics
- RD Chapter 33- Probability

RD Chapter 1- Sets |
RD Chapter 2- Relations |
RD Chapter 3- Functions |
RD Chapter 4- Measurement of Angles |
RD Chapter 5- Trigonometric Functions |
RD Chapter 6- Graphs of Trigonometric Functions |
RD Chapter 7- Trigonometric Ratios of Compound Angles |
RD Chapter 8- Transformation Formulae |
RD Chapter 9- Trigonometric Ratios of Multiple and Submultiple Angles |
RD Chapter 10- Sine and Cosine Formulae and Their Applications |
RD Chapter 11- Trigonometric Equations |
RD Chapter 12- Mathematical Induction |
RD Chapter 13- Complex Numbers |
RD Chapter 14- Quadratic Equations |
RD Chapter 15- Linear Inequations |
RD Chapter 16- Permutations |
RD Chapter 17- Combinations |
RD Chapter 18- Binomial Theorem |
RD Chapter 19- Arithmetic Progressions |
RD Chapter 20- Geometric Progressions |
RD Chapter 21- Some Special Series |
RD Chapter 22- Brief Review of Cartesian System of Rectangular Coordinates |
RD Chapter 23- The Straight Lines |
RD Chapter 24- The Circle |
RD Chapter 26- Ellipse |
RD Chapter 27- Hyperbola |
RD Chapter 28- Introduction to 3D coordinate geometry |
RD Chapter 29- Limits |
RD Chapter 30- Derivatives |
RD Chapter 31- Mathematical Reasoning |
RD Chapter 32- Statistics |
RD Chapter 33- Probability |

Find the equation of the parabola whose:

(i) focus is (3, 0) and the directrix is 3x + 4y = 1

(ii) focus is (1, 1) and the directrix is x + y + 1 = 0

(iii) focus is (0, 0) and the directrix is 2x – y – 1 = 0

(iv) focus is (2, 3) and the directrix is x – 4y + 1 = 0

**Answer
1** :

(i) focus is (3, 0) and the directrix is 3x + 4y = 1

Given:

The focus S(3, 0) and directrix(M) 3x + 4y – 1 = 0.

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Upon cross multiplication, we get

25x^{2} + 25y^{2} – 150x + 225 =9x^{2} + 16y^{2} – 6x – 8y + 24xy + 1

16x^{2} + 9y^{2} – 24xy – 144x +8y + 224 = 0

∴The equation of the parabola is 16x^{2} + 9y^{2} –24xy – 144x + 8y + 224 = 0

(ii) focus is (1, 1) and the directrix is x + y + 1 = 0

Given:

The focus S(1, 1) and directrix(M) x + y + 1 = 0.

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Upon cross multiplication, we get

2x^{2} + 2y^{2} – 4x – 4y + 4 = x^{2} +y^{2} + 2x + 2y + 2xy + 1

x^{2} + y^{2} + 2xy – 6x – 6y + 3= 0

∴ The equation of the parabola is x^{2} + y^{2} +2xy – 6x – 6y + 3 = 0

(iii) focus is (0, 0) and the directrix is 2x – y – 1 = 0

Given:

The focus S(0, 0) and directrix(M) 2x – y – 1 = 0.

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Upon cross multiplication, we get

5x^{2} + 5y^{2} = 4x^{2} +y^{2} – 4x + 2y – 4xy + 1

x^{2} + 4y^{2} + 4xy + 4x – 2y – 1= 0

∴ The equation of the parabola is x^{2} + 4y^{2} +4xy + 4x – 2y – 1 = 0

(iv) focus is (2, 3) and the directrix is x – 4y + 1 = 0

Given:

The focus S(2, 3) and directrix(M) x – 4y + 3 = 0.

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Upon cross multiplication, we get

17x^{2} + 17y^{2} – 68x – 102y +221 = x^{2} + 16y^{2} + 6x – 24y – 8xy + 9

16x^{2} + y^{2} + 8xy – 74x – 78y+ 212 = 0

∴ The equation of the parabola is 16x^{2} + y^{2} +8xy – 74x – 78y + 212 = 0

**Answer
2** :

Given:

The focus S(2, 3) and directrix(M) x – 4y + 3 = 0.

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Upon cross multiplication, we get

17x^{2} + 17y^{2} – 68x – 102y +221 = x^{2} + 16y^{2} + 6x – 24y – 8xy + 9

16x^{2} + y^{2} + 8xy – 74x – 78y+ 212 = 0

∴ The equation of the parabola is 16x^{2} + y^{2} +8xy – 74x – 78y + 212 = 0.

Now, let us find the length of the latus rectum,

We know that the length of the latus rectum is twice theperpendicular distance from the focus to the directrix.

So by using the formula,

∴ The length of the latus rectum is 14/√17

Find the equation of the parabola, if

(i) the focus is at (-6, 6) and the vertex is at (-2, 2)

(ii) the focus is at (0, -3) and the vertex is at (0, 0)

(iii) the focus is at (0, -3) and the vertex is at (-1, -3)

(iv) the focus is at (a, 0) and the vertex is at (a′, 0)

(v) the focus is at (0, 0) and the vertex is at the intersection of the lines x + y = 1 and x – y = 3.

**Answer
3** :

(i) the focus is at (-6, 6) and the vertex is at (-2, 2)

Given:

Focus = (-6, 6)

Vertex = (-2, 2)

Let us find the slope of the axis (m_{1}) =(6-2)/(-6-(-2))

= 4/-4

= -1

Let us assume m_{2} be the slope of thedirectrix.

m_{1}m_{2} = -1

-1m_{2} = -1

m_{2} = 1

Now, let us find the point on directrix.

(-2, 2) = [(x-6/2), (y+6)/2]

By equating we get,

(x-6/2) = -2 and (y+6)/2 = 2

x-6 = -4 and y+6 = 4

x = -4+6 and y = 4-6

x = 2 and y = -2

So the point of directrix is (2, -2).

We know that the equation of the lines passing through (x_{1},y_{1}) and having slope m is y – y_{1} = m(x – x_{1})

y – (-2) = 1(x – 2)

y + 2 = x – 2

x – y – 4 = 0

Let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Now by cross multiplying, we get

2x^{2} + 2y^{2} + 24x – 24y + 144= x^{2} + y^{2} – 8x + 8y – 2xy + 16

x^{2} + y^{2} + 2xy + 32x – 32y +128 = 0

∴ The equation of the parabola is x^{2} + y^{2} +2xy + 32x – 32y + 128 = 0

(ii) the focus is at (0, -3) and the vertex is at (0, 0)

Given:

Focus = (0, -3)

Vertex = (0, 0)

Let us find the slope of the axis (m_{1}) =(-3-0)/(0-0)

= -3/0

= ∞

Since the axis is parallel to the x-axis, the slope of thedirectrix is equal to the slope of x-axis = 0

So, m_{2} = 0

Now, let us find the point on directrix.

(0, 0) = [(x-0/2), (y-3)/2]

By equating we get,

(x/2) = 0 and (y-3)/2 = 0

x = 0 and y – 3 = 0

x = 0 and y = 3

So the point on directrix is (0, 3).

We know that the equation of the lines passing through (x_{1},y_{1}) and having slope m is y – y_{1} = m(x – x_{1})

y – 3 = 0(x – 0)

y – 3 = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

Now by cross multiplying, we get

x^{2} + y^{2} + 6y + 9 = y^{2} –6y + 9

x^{2} + 12y = 0

∴ The equation of the parabola is x^{2} + 12y =0

(iii) the focus is at (0, -3) and the vertex is at (-1, -3)

Given:

Focus = (0, -3)

Vertex = (-1, -3)

Let us find the slope of the axis (m_{1}) =(-3-(-3))/(0-(-1))

= 0/1

= 0

We know, the products of the slopes of the perpendicularlines is – 1 for non – vertical lines.

Here the slope of the axis is equal to the slope of the x –axis. So, the slope of directrix is equal to the slope of y – axis i.e., ∞.

So, m_{2} = ∞

Now let us find the point on directrix.

(-1, -3) = [(x+0/2), (y-3)/2]

By equating we get,

(x/2) = -1 and (y-3)/2 = -3

x = -2 and y – 3 = -6

x = -2 and y = -6+3

x = -2 and y = -3

So, the point on directrix is (-2, -3)

We know that the equation of the lines passing through (x_{1},y_{1}) and having slope m is y – y_{1} = m(x – x_{1})

y – (- 3) = ∞(x – (- 2))

(y+3)/ ∞ = x + 2

x + 2 = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

By cross multiplying, we get

x^{2} + y^{2} + 6y + 9 = x^{2} +4x + 4

y^{2} – 4x + 6y + 5 = 0

∴ The equation of the parabola is y^{2} – 4x +6y + 5 = 0

(iv) the focus is at (a, 0) and the vertex is at (a′, 0)

Given:

Focus = (a, 0)

Vertex = (a′, 0)

Let us find the slope of the axis (m_{1}) =(0-0)/(a′, a)

= 0/(a′, a)

= 0

We know, the products of the slopes of the perpendicularlines is – 1 for non – vertical lines.

Here the slope of the axis is equal to the slope of the x –axis. So, the slope of directrix is equal to the slope of y – axis i.e., ∞.

So, m_{2} = ∞

Now let us find the point on directrix.

(a′, 0)= [(x+a/2), (y+0)/2]

By equating we get,

(x+a/2) = a′ and (y)/2 = 0

x + a = 2a′ and y = 0

x = (2a′ – a) and y = 0

So the point on directrix is (2a′ – a, 0).

_{1},y_{1}) and having slope m is y – y_{1} = m(x – x_{1})

y – (0) = ∞(x – (2a′ – a))

y/∞ = x + a – 2a′

x + a – 2a′ = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

By cross multiplying we get,

x^{2} + y^{2} – 2ax + a^{2} =x^{2} + a^{2} + 4(a′)^{2} + 2ax – 4aa′ –4a′x

y^{2} – (4a – 4a′)x + a^{2} –4(a′)^{2} + 4aa′ = 0

∴ The equation of the parabola is y^{2} – (4a –4a′)x + a^{2} – 4(a′)^{2} + 4aa′ = 0

(v) the focus is at (0, 0) and the vertex is at the intersectionof the lines x + y = 1 and x – y = 3.

Given:

Focus = (0, 0)

Vertex = intersection of the lines x + y = 1 and x – y = 3.

So the intersecting point of above lines is (2, -1)

Vertex = (2, -1)

Slope of axis (m_{1}) = (-1-0)/(2-0)

= -1/2

We know that the products of the slopes of the perpendicularlines is – 1.

Let us assume m_{2} be the slope of thedirectrix.

m_{1}.m_{2} = -1

-1/2 . m_{2} = -1

So m_{2} = 2

Now let us find the point on directrix.

(2, -1) = [(x+0)/2, (y+0)/2]

(x)/2 = 2 and y/2 = -1

x = 4 and y = -2

The point on directrix is (4, – 2).

_{1},y_{1}) and having slope m is y – y_{1} = m(x – x_{1})

y – (- 2) = 2(x – 4)

y + 2 = 2x – 8

2x – y – 10 = 0

Now, let us assume P(x, y) be any point on the parabola.

The distance between two points (x_{1}, y_{1})and (x_{2}, y_{2}) is given as:

And the perpendicular distance from the point (x_{1},y_{1}) to the line ax + by + c = 0 is

So by equating both, we get

By cross multiplying, we get

5x^{2} + 5y^{2} = 4x^{2} +y^{2} – 40x + 20y – 4xy + 100

x^{2} + 4y^{2} + 4xy + 40x – 20y –100 = 0

∴ The equation of the parabola is x^{2} + 4y^{2} +4xy + 40x – 20y – 100 = 0

Find the vertex, focus, axis,directrix and lotus – rectum of the following parabolas

(i) y^{2} = 8x

(ii) 4x^{2} + y = 0

(iii) y^{2} – 4y – 3x+ 1 = 0

(iv) y^{2} – 4y + 4x= 0

(v) y^{2} + 4x + 4y –3 = 0

**Answer
4** :

(i) y^{2} = 8x

Given:

Parabola = y^{2} = 8x

Now by comparing with the actual parabola y^{2} =4ax

Then,

4a = 8

a = 8/4 = 2

So, the vertex is (0, 0)

The focus is (a, 0) = (2, 0)

The equation of the axis is y = 0.

The equation of the directrix is x = – a i.e., x = – 2

The length of the latus rectum is 4a = 8.

(ii) 4x^{2} + y = 0

Given:

Parabola => 4x^{2} + y = 0

Now by comparing with the actual parabola y^{2} =4ax

Then,

4a = ¼

a = 1/(4 × 4)

= 1/16

So, the vertex is (0, 0)

The focus is = (0, -1/16)

The equation of the axis is x = 0.

The equation of the directrix is y = 1/16

The length of the latus rectum is 4a = ¼

(iii) y^{2} – 4y – 3x + 1 = 0

Given:

Parabola y^{2} – 4y – 3x + 1 = 0

y^{2} – 4y = 3x – 1

y^{2} – 4y + 4 = 3x + 3

(y – 2)^{2} = 3(x + 1)

Now by comparing with the actual parabola y^{2} =4ax

Then,

4b = 3

b = ¾

So, the vertex is (-1, 2)

The focus is = (3/4 – 1, 2) = (-1/4, 2)

The equation of the axis is y – 2 = 0.

The equation of the directrix is (x – c) = -b

(x – (-1)) = -3/4

x = -1 – ¾

= -7/4

The length of the latus rectum is 4b = 3

(iv) y^{2} – 4y + 4x = 0

Given:

Parabola y^{2} – 4y + 4x = 0

y^{2} – 4y = – 4x

y^{2} – 4y + 4 = – 4x + 4

(y – 2)^{2} = – 4(x – 1)

Now by comparing with the actual parabola y^{2} =4ax => (y – a)^{2} = – 4b(x – c)

Then,

4b = 4

b = 1

So, the vertex is (c, a) = (1, 2)

The focus is (b + c, a) = (1-1, 2) = (0, 2)

The equation of the axis is y – a = 0 i.e., y – 2 = 0

The equation of the directrix is x – c = b

x – 1 = 1

x = 1 + 1

= 2

Length of latus rectum is 4b = 4

(v) y^{2} + 4x + 4y – 3 = 0

Given:

The parabola y^{2} + 4x + 4y – 3 = 0

y^{2} + 4y = – 4x + 3

y^{2} + 4y + 4 = – 4x + 7

(y + 2)^{2} = – 4(x – 7/4)

Now by comparing with the actual parabola y^{2} =4ax => (y – a)^{2} = – 4b(x – c)

Then,

4b = 4

b = 4/4 = 1

So, The vertex is (c, a) = (7/4, -2)

The focus is (- b + c, a) = (-1 + 7/4, -2) = (3/4, -2)

The equation of the axis is y – a = 0 i.e., y + 2 = 0

The equation of the directrix is x – c = b

x – 7/4 = 1

x = 1 + 7/4

= 11/4

Length of latus rectum is 4b = 4.

For the parabola, y^{2} =4px find the extremities of a double ordinate of length 8p. Prove that thelines from the vertex to its extremities are at right angles.

**Answer
5** :

Given:

The parabola, y^{2} = 4px and a double ordinateof length 8p.

Let AB be the double ordinate of length 8p for the parabolay^{2} = 4px.

Now, let us compare to the actual parabola, y^{2} =4ax

Then,

axis is y = 0

vertex is O(0, 0).

We know that double ordinate is perpendicular to the axis.

So, let us assume that the point at which the doubleordinate meets the axis is (x_{1}, 0).

Then the equation of the double ordinate is y = x_{1}.It meets the parabola at the points (x_{1}, 4p) and (x_{1}, –4p) as its length is 8p.

Now, let us find the value of x_{1} bysubstituting in the parabola.

(4p)^{2} = 4p(x_{1})

x_{1} = 4p.

The extremities of the double ordinate are A(4p, 4p) andB(4p, – 4p).

Assume the slopes of OA and OB be m_{1} and m_{2}.Let us find their values.

m_{1} = (4p – 0)/(4p – 0)

= 4p/4p

= 1

m_{2} = (4p – 0)/(-4p – 0)

= 4p/-4p

= -1

So, m_{1}.m_{2} = 1. – 1

= – 1

The product of slopes is -1. So, the lines OA and OB areperpendicular to each other.

Hence the extremities of double ordinate make right anglewith the vertex.

**Answer
6** :

Given:

The parabola, x^{2} = 12y

Now, let us compare to the actual parabola, y^{2} =4ax

Then,

Vertex is O(0, 0)

Ends of latus rectum is (2b, b), (-2b, b)

4b = 12

b = 12/4

= 3

Ends of latus rectum = (2(3), 3), (- 2(3), 3)

Ends of latus rectum is A(6, 3), B(- 6, 3)

We know that area of the triangle with the vertices (x_{1},y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})is

∴The area of the triangle is 18 sq.units.

**Answer
7** :

Given:

Focus = (3, 3)

Directrix = 3x – 4y = 2

Firstly let us find the slope of the directrix.

The slope of the line ax + by + c = 0 is –a/b

So, m_{1} = -3/-4

= ¾

Let us assume the slope of axis is m_{2}.

m_{1}.m_{2} = -1

¾ . m_{2} = -1

m_{2} = -4/3

We know that the equation of the line passing through thepoint (x_{1}, y_{1}) and having slope m is (y – y_{1})= m(x – x_{1})

y – 3 = -4/3 (x – 3)

3(y – 3) = – 4(x – 3)

3y – 9 = – 4x + 12

4x + 3y = 21

On solving the lines, the intersection point is (18/5, 11/5)

By using the formula to find the length is given as

∴The length of the latus rectum is 2.

**Answer
8** :

Given:

The parabola, x^{2} = 9y

Let us assume the point be (3y_{1}, y_{1}).

Now by substituting the point in the parabola we get,

(3y_{1})^{2} = 9(y_{1})

9y_{1}^{2} = 9y_{1}

y_{1}^{2} – y_{1} = 0

y_{1}(y_{1} – 1) = 0

y_{1} = 0 or y_{1} – 1 = 0

y_{1} = 0 or y_{1} = 1

The points is B (3(1), 1) => (3, 1)

∴The point is (3, 1).

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