If P and Q are two points whose coordinates are `(a t^2,2a t)a n d(a/(t^2),(2a)/t)` respectively and S is the point (a,0). Show that `1/(S P)+1/(s Q)` is independent of t.

Find the locus of a point whose coordinate are given by x = t+t^(2), y = 2t+1 , where t is variable.

Two particles P and Q are moving with velocity of (hat(i)+hat(j)) and (-hat(i)+2hat(j)) respectively. At time t=0 , P is at origin and Q is at a point with positive vector (2hat(i)+hat(j)) . Then the shortest distance between P & Q is :-

If the coordinates of a vartiable point P be (t+(1)/(t), t-(1)/(t)) , where t is the variable quantity, then the locus of P is

Football teams T_1 and T_2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T_1 winning. Drawing and losing a game against T_2 are (1)/(2),(1)/(6) and (1)/(3) respectively. Each team gets 3 points for a win. 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T_1 and T_2 respectively. after two games.P(X>Y) is

Football teams T_1 and T_2 have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of T_1 winning, drawing and losing a game against T_2 are (1)/(2),(1)/(6) and (1)/(3) respectively. Each team gets 3 points for a win. 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by teams T_1 and T_2 respectively. after two games. P(X=Y) is

Find all points of dicontinuity of the function f(t)= (1)/(t^(2) + t-2) , where t = (1)/(x-1)

Write the equation of a tangent to the curve x=t, y=t^2 and z=t^3 at its point M(1, 1, 1): (t=1) .

The position of a particle is given by r=3.0thati+2.0t^(2)hatj+5,0hatk where t is in seconds and the coefficients have the proper units for r to be in matres. (a) Find v (t) and a(t) of the particle . (b) Find the magnitude and direction of v (t) at t=1.0 s.

The displacement of particle with respect to time is s = 3^(t3) - 7t^(2) + 5t + 8 where s is in m and t is in s, then acceleration of particle at t = ls is

The slope of the tangent to the curve x=t^(2)+3t-8, y=2t^(2)-2t-5 at the point (2,-1) is …………