Maths-

General

Easy

Question

# The polar equation of is

Hint:

### The term "polar coordinate system" refers to a two-dimensional coordinate system where each point's location on a plane is determined by its distance from a reference point and its angle with respect to a reference direction. Here we have to find the polar equation of .

## The correct answer is:

### Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as:

Now lets xonsider: x=rcosθ and y=rsinθ

x^{2}+y^{2}=r^{2}

Now putting the values, we get:

Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. So the equation is .

### Related Questions to study

Maths-

### The cartesian equation of is

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as

### The cartesian equation of is

Maths-General

Using the formula, we may generate an endless number of polar coordinates for a single coordinate point.

(r, +2n) or (-r, +(2n+1)) are possible formulas, where n is an integer.

If measured in the opposite direction, the value of is positive.

When calculated counterclockwise, the value of is negative.

If laid off at the terminal side of, the value of r is positive.

If r is terminated at the prolongation through the origin from the terminal side of, it has a negative value.

The equation is given as

physics-

### Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be

or

On waxing the number of beats decreases hence

On waxing the number of beats decreases hence

### Two tuning forks and are vibrated together. The number of beats produced are represented by the straight line in the following graph. After loading with wax again these are vibrated together and the beats produced are represented by the line If the frequency of is the frequency of will be

physics-General

or

On waxing the number of beats decreases hence

On waxing the number of beats decreases hence

Maths-

### If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are

a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

### If a hyperbola passing through the origin has and as its asymptotes, then the equation of its tranvsverse and conjugate axes are

Maths-General

a line or curve that serves as the boundary of another line or curve in mathematics. An example of an asymptotic curve is a descending curve that approaches but does not reach the horizontal axis, which is the asymptote of the curve.

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

The transverse axis is the bisector containing origin, and the hyperbola's axes are the bisectors of the pair of asmptodes.

So we have:

maths-

### Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

### Five distinct letters are to be transmitted through a communication channel. A total number of 15 blanks is to be inserted between the two letters with at least three between every two. The number of ways in which this can be done is -

maths-General

For 1 i 4, let x

x

The number of solutions of (1)

= coefficient of t

= coefficient of t

= coefficient of t

=

But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

_{i}( 3) be the number of blanks between i^{th}and (i + 1)^{th}letters. Then,x

_{1}+ x_{2}+ x_{3}+ x_{4}= 15 …..(1)The number of solutions of (1)

= coefficient of t

^{15}in (t^{3}+ t^{4}+….)^{4}= coefficient of t

^{3}in (1 – t)^{–4}= coefficient of t

^{3}in [1 +^{4}C_{1}+^{5}C_{2}t^{2}+^{6}C_{3 }t^{3}+ …..]=

^{6}C_{3}= 20.But 5 letters can be permuted in 5! = 120 ways.

Thus, the required number arrangements

= (120) (20) = 2400.

Hence

maths-

### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

maths-General

Note that 7

Thus, 7

For 7

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

Hence

^{r}(r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).Thus, 7

^{m}+ 7^{n}cannot end in 5 for any values of m, n N. In other words, for 7^{m}+ 7^{n}to be divisible by 5, it should end in 0.For 7

^{m}+ 7^{n}to end in 0, the forms of m and n should be as follows :m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7

^{m}+ 7^{n}ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98] There are 100 × 25 = 2500 ordered pairs (m, n) for which 7

^{m}+ 7^{n}is divisible by 5.Hence

maths-

### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

maths-General

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

maths-

### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

maths-General

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

maths-

### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

maths-General

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).maths-

### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

maths-General

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.maths-

### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

maths-General

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.Maths-

### If the line is a normal to the hyperbola then

### If the line is a normal to the hyperbola then

Maths-General

Maths-

### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.

We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.

We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.

### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

Maths-General

A hyperbola is a significant conic section in mathematics that is created by the intersection of a double cone with a plane surface, though not always at the centre. A hyperbola is symmetric along its conjugate axis and resembles the ellipse in many ways. A hyperbola is subject to concepts like foci, directrix, latus rectus, and eccentricity.

We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.

We have given: the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola.

Maths-

### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

Maths-General

Maths-

### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.

Now we have given the LCM as: r

Consider following cases:

Case 1: if p contains r

So 2 ways.

Case 2: if q contains r

So 2 ways.

Case 3:Both p and q contain r

So 1way.

So after this we can say that:

exponent of r=2+2+1 = 5 ways.

Similarly

exponent of t=4+4+1=9 ways.

exponent of s=2+2+1 = 5 ways.

So total ways will be:

5 x 5 x 9 = 225 ways.

Now we have given the LCM as: r

^{2}t^{4}s^{2}Consider following cases:

Case 1: if p contains r

^{2}then q will have r^{k}, for the value k=0,1.So 2 ways.

Case 2: if q contains r

^{2}then p will have r^{k}, for the value k=0,1.So 2 ways.

Case 3:Both p and q contain r

^{2}So 1way.

So after this we can say that:

exponent of r=2+2+1 = 5 ways.

Similarly

exponent of t=4+4+1=9 ways.

exponent of s=2+2+1 = 5 ways.

So total ways will be:

5 x 5 x 9 = 225 ways.

### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

Maths-General

An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.

Now we have given the LCM as: r

Consider following cases:

Case 1: if p contains r

So 2 ways.

Case 2: if q contains r

So 2 ways.

Case 3:Both p and q contain r

So 1way.

So after this we can say that:

exponent of r=2+2+1 = 5 ways.

Similarly

exponent of t=4+4+1=9 ways.

exponent of s=2+2+1 = 5 ways.

So total ways will be:

5 x 5 x 9 = 225 ways.

Now we have given the LCM as: r

^{2}t^{4}s^{2}Consider following cases:

Case 1: if p contains r

^{2}then q will have r^{k}, for the value k=0,1.So 2 ways.

Case 2: if q contains r

^{2}then p will have r^{k}, for the value k=0,1.So 2 ways.

Case 3:Both p and q contain r

^{2}So 1way.

So after this we can say that:

exponent of r=2+2+1 = 5 ways.

Similarly

exponent of t=4+4+1=9 ways.

exponent of s=2+2+1 = 5 ways.

So total ways will be:

5 x 5 x 9 = 225 ways.

Maths-

### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.

There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).

Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.

Then the number of rectangles will be:

(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m

So it is m

There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).

Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.

Then the number of rectangles will be:

(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m

^{2}n^{2}So it is m

^{2}n^{2}.### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

Maths-General

Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.

There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).

Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.

Then the number of rectangles will be:

(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m

So it is m

There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).

Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.

Then the number of rectangles will be:

(1+3+5+......+(2m−1))(1+3+5+......+(2n−1))=m

^{2}n^{2}So it is m

^{2}n^{2}.