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
x2+y2=r2
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
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 henceMaths-
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 xi (
3) be the number of blanks between ith and (i + 1)th letters. Then,
x1 + x2 + x3 + x4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t15 in (t3 + t4 +….)4
= coefficient of t3 in (1 – t)–4
= coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
= 6C3 = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
i 4, let xi ( 3) be the number of blanks between ith and (i + 1)th letters. Then,x1 + x2 + x3 + x4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t15 in (t3 + t4 +….)4
= coefficient of t3 in (1 – t)–4
= coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
= 6C3 = 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 xi ( 3) be the number of blanks between ith and (i + 1)th letters. Then,
x1 + x2 + x3 + x4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t15 in (t3 + t4 +….)4
= coefficient of t3 in (1 – t)–4
= coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
= 6C3 = 20.
But 5 letters can be permuted in 5! = 120 ways.
Thus, the required number arrangements
= (120) (20) = 2400.
Hence
i 4, let xi ( 3) be the number of blanks between ith and (i + 1)th letters. Then,x1 + x2 + x3 + x4 = 15 …..(1)
The number of solutions of (1)
= coefficient of t15 in (t3 + t4 +….)4
= coefficient of t3 in (1 – t)–4
= coefficient of t3 in [1 + 4C1 + 5C2 t2 + 6C3 t3 + …..]
= 6C3 = 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 7m + 7n is divisible by 5 is -
Note that 7r (r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
Thus, 7m + 7n cannot end in 5 for any values of m, n N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
For 7m + 7n 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 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
Hence
Thus, 7m + 7n cannot end in 5 for any values of m, n N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
For 7m + 7n 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 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
Hence
The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7m + 7n is divisible by 5 is -
maths-General
Note that 7r (r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
Thus, 7m + 7n cannot end in 5 for any values of m, n N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
For 7m + 7n 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 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
Hence
Thus, 7m + 7n cannot end in 5 for any values of m, n N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
For 7m + 7n 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 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n 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
=
= =
=
= 35Hence (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
=
= =
=
= 35Hence (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)2 – 2[k + (k – 1) + …. + 2 + 1]
= (3k2 + 3k + 1).
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 + kNumber of points having integral coordinates= (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
= (3k2 + 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)2 – 2[k + (k – 1) + …. + 2 + 1]
= (3k2 + 3k + 1).
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 + kNumber of points having integral coordinates= (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
= (3k2 + 3k + 1).
maths-
The numbers of integers between 1 and 106 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 (x1 + x2 + x3 + x4 + x5 + x6 = K)
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
The numbers of integers between 1 and 106 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 (x1 + x2 + x3 + x4 + x5 + x6 = K)
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
= Coefficient of xK in (1 + x + x2 +….. + x9)6
= Coefficient of xK in
= Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
(1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
= k + 6CK – 6. K–4CK–10
= k + 6C6 – 6. K–4C6 .
maths-
The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. 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 + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
m + n + kC3 – mC3 – nC3 – kC3.
m + n + kC3 – mC3 – nC3 – kC3.
The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. 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 + kC3. But out of these m + n + k points, m points lie on I1, n points lie on I2 and k points lie on I3 and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is
m + n + kC3 – mC3 – nC3 – kC3.
m + n + kC3 – mC3 – nC3 – kC3.
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 r2t4s2, 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: r2t4s2
Consider following cases:
Case 1: if p contains r2 then q will have rk, for the value k=0,1.
So 2 ways.
Case 2: if q contains r2 then p will have rk, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r2
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: r2t4s2
Consider following cases:
Case 1: if p contains r2 then q will have rk, for the value k=0,1.
So 2 ways.
Case 2: if q contains r2 then p will have rk, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r2
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 r2t4s2, 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: r2t4s2
Consider following cases:
Case 1: if p contains r2 then q will have rk, for the value k=0,1.
So 2 ways.
Case 2: if q contains r2 then p will have rk, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r2
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: r2t4s2
Consider following cases:
Case 1: if p contains r2 then q will have rk, for the value k=0,1.
So 2 ways.
Case 2: if q contains r2 then p will have rk, for the value k=0,1.
So 2 ways.
Case 3:Both p and q contain r2
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))=m2n2
So it is m2n2.
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))=m2n2
So it is m2n2.
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))=m2n2
So it is m2n2.
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))=m2n2
So it is m2n2.
