Maths-
General
Easy

Question

The equation of the line passing through pole and open parentheses 2 comma fraction numerator pi over denominator 3 end fraction close parentheses is

  1. r=2    
  2. theta equals pi divided by 3    
  3. r to the power of 2 end exponent equals 2 r plus 3    
  4. r S i n space theta equals 4    

The correct answer is: theta equals pi divided by 3

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Related Questions to study

General
Maths-

The polar equation of y to the power of 2 end exponent equals 4 x 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: y to the power of 2 end exponent equals 4 x
Now lets xonsider: x=rcosθ and y=rsinθ
x2+y2=r2
Now putting the values, we get:
y squared equals 4 x
r squared sin squared theta equals 4 r cos theta
r sin squared theta equals 4 cos theta
r equals 4 fraction numerator cos theta over denominator sin squared theta end fraction
r equals 4 space c o t space theta space cos e c space theta

The polar equation of y to the power of 2 end exponent equals 4 x 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: y to the power of 2 end exponent equals 4 x
Now lets xonsider: x=rcosθ and y=rsinθ
x2+y2=r2
Now putting the values, we get:
y squared equals 4 x
r squared sin squared theta equals 4 r cos theta
r sin squared theta equals 4 cos theta
r equals 4 fraction numerator cos theta over denominator sin squared theta end fraction
r equals 4 space c o t space theta space cos e c space theta
General
Maths-

The cartesian equation of r equals a s i n space 2 theta 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
r equals a sin space 2 theta
N o w space w e space k n o w space t h a t colon
sin space 2 theta equals 2 s i n theta cos theta
A p p l y i n g space t h i s comma space w e space g e t colon
r equals 2 a s i n theta cos theta
M u l t i p l y i n g space b y space r squared space o n space b o t h space s i d e s comma space w e space g e t colon
r cubed equals 2 a left parenthesis r s i n theta right parenthesis left parenthesis r cos theta right parenthesis
P u t t i n g space r cos theta equals x space a n d space space r s i n theta equals y comma space w e space g e t colon
r equals square root of x squared plus y squared end root
left parenthesis x squared plus y squared right parenthesis to the power of 3 divided by 2 end exponent equals 2 a y x space
S q u a r i n g space b o t h space s i d e s comma space w e space g e t colon
4 a squared x squared y squared equals left parenthesis x squared plus y squared right parenthesis cubed

The cartesian equation of r equals a s i n space 2 theta 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
r equals a sin space 2 theta
N o w space w e space k n o w space t h a t colon
sin space 2 theta equals 2 s i n theta cos theta
A p p l y i n g space t h i s comma space w e space g e t colon
r equals 2 a s i n theta cos theta
M u l t i p l y i n g space b y space r squared space o n space b o t h space s i d e s comma space w e space g e t colon
r cubed equals 2 a left parenthesis r s i n theta right parenthesis left parenthesis r cos theta right parenthesis
P u t t i n g space r cos theta equals x space a n d space space r s i n theta equals y comma space w e space g e t colon
r equals square root of x squared plus y squared end root
left parenthesis x squared plus y squared right parenthesis to the power of 3 divided by 2 end exponent equals 2 a y x space
S q u a r i n g space b o t h space s i d e s comma space w e space g e t colon
4 a squared x squared y squared equals left parenthesis x squared plus y squared right parenthesis cubed
General
physics-

Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
On waxing Q comma the number of beats decreases hence
n subscript Q end subscript equals 344 H z

Two tuning forks P and Q are vibrated together. The number of beats produced are represented by the straight line O A in the following graph. After loading Q with wax again these are vibrated together and the beats produced are represented by the line O B. If the frequency of P is 341 H z comma the frequency of Q will be

physics-General
n subscript Q end subscript equals 341 plus-or-minus 3 equals 344 H z or 338 H z
On waxing Q comma the number of beats decreases hence
n subscript Q end subscript equals 344 H z
General
Maths-

If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 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:
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals 4 x minus 3 y minus 6
3 x minus 4 y minus 1 minus 4 x plus 3 y plus 6 equals 0
x plus y minus 5 equals 0
T h i s space i s space t h e space e q u a t i o n space o f space t r a n s v e r s e.
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals negative space 4 x plus 3 y plus space 6
3 x minus 4 y minus 1 plus space 4 x minus 3 y minus space 6 equals 0
7 x minus 7 y equals 7
x minus y minus 1 equals 0
T h i s space i s space e q u a t i o n space o f space c o n j u g a t e space a x i s.

If a hyperbola passing through the origin has 3 x minus 4 y minus 1 equals 0 and 4 x minus 3 y minus 6 equals 0 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:
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals plus space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals 4 x minus 3 y minus 6
3 x minus 4 y minus 1 minus 4 x plus 3 y plus 6 equals 0
x plus y minus 5 equals 0
T h i s space i s space t h e space e q u a t i o n space o f space t r a n s v e r s e.
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 3 squared plus left parenthesis negative 4 right parenthesis squared end root end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 4 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction
fraction numerator 3 x minus 4 y minus 1 over denominator square root of 25 end fraction equals negative space fraction numerator 4 x minus 3 y minus 6 over denominator square root of 25 end fraction
3 x minus 4 y minus 1 equals negative space 4 x plus 3 y plus space 6
3 x minus 4 y minus 1 plus space 4 x minus 3 y minus space 6 equals 0
7 x minus 7 y equals 7
x minus y minus 1 equals 0
T h i s space i s space e q u a t i o n space o f space c o n j u g a t e space a x i s.
General
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 less or equal thanless or equal than 4, let xi (greater or equal than 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 less or equal thanless or equal than 4, let xi (greater or equal than 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
General
maths-

The number of ordered pairs (m, n), m, n element of {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

The number of ordered pairs (m, n), m, n element of {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
General
maths-

Consider the following statements:
1. The number of ways of arranging m different things taken all at a time in which p less or equal than 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 fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
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 = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
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 less or equal than 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 fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
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 = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
Hence, both of statements are correct.
General
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) less or equal than ƒ(j), straight for all i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,
element of {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
= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
= fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
= fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 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) less or equal than ƒ(j), straight for all i < j, is equal to-

maths-General
Let ‘l’ is associated with ‘r’ ,
element of {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
= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
= fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
= fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 35
Hence (a) is correct answer.
General
maths-

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

|x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
& |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
& |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
thereforeNumber 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|  less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

maths-General
|x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
& |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
& |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
thereforeNumber of points having integral coordinates
= (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
= (3k2 + 3k + 1).
General
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 open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
= 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 open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
= 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 .
General
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 + kC3mC3nC3kC3.

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 + kC3mC3nC3kC3.
General
Maths-

If the line l x plus m y equals 1 is a normal to the hyperbola fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 then fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals

If the line l x plus m y equals 1 is a normal to the hyperbola fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 then fraction numerator a to the power of 2 end exponent over denominator l to the power of 2 end exponent end fraction minus fraction numerator b to the power of 2 end exponent over denominator m to the power of 2 end exponent end fraction equals

Maths-General
General
Maths-

If the tangents drawn from a point on the hyperbola x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 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 x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 make angles α and β with the transverse axis of the hyperbola. 
N o w space l e t s space t a k e space a space p o i n t space P space o n space t h e space h y p e r b o l a space w i t h space c o o r d i n a t e s space left parenthesis h comma k right parenthesis comma space t h e n colon
h squared minus k squared equals a squared minus b squared
N o w space t h e space e q u a t i o n space o f space tan g e n t space t o space t h e space e l l i p s e space x squared over a squared plus y squared over b squared equals 1 space w i l l space b e colon
y equals m x plus square root of a squared m squared plus b squared end root
S i m p l i f y i n g space i t comma space w e space g e t colon
left parenthesis k minus m h right parenthesis squared equals a to the power of 2 end exponent m to the power of 2 end exponent plus b squared
m squared left parenthesis h squared minus a squared right parenthesis minus 2 m h k plus k squared minus b squared equals 0
N o w space l e t space m 1 space a n d space m 2 space b e space t h e space r o o t space o f space t h e space e q u a t i o n space f o r m e d comma space t h e n space w e space g e t colon
m 1 m 2 equals fraction numerator k squared minus b squared over denominator h squared minus a squared end fraction equals 1
N o w space u sin g space e q u a t i o n space 1 comma space w e space g e t colon
tan alpha space tan beta equals 1

If the tangents drawn from a point on the hyperbola x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 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 x to the power of 2 end exponent minus y to the power of 2 end exponent equals a to the power of 2 end exponent minus b to the power of 2 end exponent to the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction minus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1 make angles α and β with the transverse axis of the hyperbola. 
N o w space l e t s space t a k e space a space p o i n t space P space o n space t h e space h y p e r b o l a space w i t h space c o o r d i n a t e s space left parenthesis h comma k right parenthesis comma space t h e n colon
h squared minus k squared equals a squared minus b squared
N o w space t h e space e q u a t i o n space o f space tan g e n t space t o space t h e space e l l i p s e space x squared over a squared plus y squared over b squared equals 1 space w i l l space b e colon
y equals m x plus square root of a squared m squared plus b squared end root
S i m p l i f y i n g space i t comma space w e space g e t colon
left parenthesis k minus m h right parenthesis squared equals a to the power of 2 end exponent m to the power of 2 end exponent plus b squared
m squared left parenthesis h squared minus a squared right parenthesis minus 2 m h k plus k squared minus b squared equals 0
N o w space l e t space m 1 space a n d space m 2 space b e space t h e space r o o t space o f space t h e space e q u a t i o n space f o r m e d comma space t h e n space w e space g e t colon
m 1 m 2 equals fraction numerator k squared minus b squared over denominator h squared minus a squared end fraction equals 1
N o w space u sin g space e q u a t i o n space 1 comma space w e space g e t colon
tan alpha space tan beta equals 1
General
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
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.

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.