The polar equation of $y to the power of 2 end exponent equals 4 x$ is

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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

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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

Here we used the concept of polar coordinate system and also the trigonometric ratios to find the solution. $T h e space e q u a t i o n space o f space t r a n s v e r s e space i s space x plus y minus 5 equals 0. space T h e space e q u a t i o n space o f space c o n j u g a t e space a x i s space i s space x minus y minus 1 equals 0.$

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

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 -

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

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 $fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction$ ways.
Which of these is/are correct?

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-

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-

The numbers of integers between 1 and 10⁶ have the sum of their digit equal to K(where 0 < K < 18) is -

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 -

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 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

So here we understood the concept of hyperbola and the normal lines.In analytic geometry, a hyperbola is a conic section created when a plane meets a double right circular cone at an angle that overlaps both cone halves. So the correct relation is $tan alpha space tan beta equals 1$

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

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

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 –

Finding the smallest common multiple between any two or more numbers is done using the least common multiple (LCM) approach. A number that is a multiple of two or more other numbers is said to be a common multiple. Here we understood the concept of LCM and the pairs, so the total pairs can be 225.

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 –

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

Here we used the concept of number system and the rectangle, we can also solve it by permutation and combination. herefore, we get the number of rectangles possible with odd side length = 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