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What will be the value of the co-ordinate whose position of a particle moving along the parabola y2 = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?
  • a)
    (1, 1)
  • b)
    (2, 2)
  • c)
    (3, 3)
  • d)
    (4, 4)
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
What will be the value of the co-ordinate whose position of a particle...
To find the coordinate of the particle, we need to determine the rates of increase of the abscissa (x-coordinate) and the ordinate (y-coordinate) along the parabola y^2 = 4x.

Let's differentiate the equation of the parabola with respect to x to find the rate of increase of y with respect to x.

Differentiating y^2 = 4x with respect to x using the chain rule, we get:
2yy' = 4
y' = 4/(2y)
y' = 2/y

Similarly, let's differentiate the equation of the parabola with respect to x to find the rate of increase of x with respect to y.

Differentiating y^2 = 4x with respect to y using the chain rule, we get:
2yy' = 4
y' = 4/(2y)
y' = 2/y

Since the rate of increase of the abscissa is twice the rate of increase of the ordinate, we have:
2/y = 2/(2/y)
2/y = y/2
4 = y^2
y = ±2

Now, substituting the value of y into the equation of the parabola, we can find the corresponding x-coordinate.

For y = 2:
y^2 = 4x
4 = 4x
x = 1

For y = -2:
y^2 = 4x
4 = 4x
x = 1

Therefore, the coordinate of the particle is (1, ±2).

Since the question asks for the coordinate where the rate of increase of the abscissa is twice the rate of increase of the ordinate, we consider the positive value of y, which is y = 2. Thus, the coordinate of the particle is (1, 2).

Hence, the correct answer is option D, (1, 2).
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Community Answer
What will be the value of the co-ordinate whose position of a particle...
Here, y2 = 4x ….(1)
Let, (x, y) be the position of the particle moving along the parabola (1) at time t.
Now, differentiating both sides of (1) with respect to t, we get:
2y(dy/dt) = 4(dx/dt)
Or, y(dy/dt) = 2(dy/dt) ……….(2)
By question, dx/dt = 2 * dy/dt ……….(3)
From (2) and (3) we get, y(dy/dt) = 2 * 2 dy/dt
Or, y = 4
Putting y = 4 in (1) we get, 42 = 4x
So, x = 4
Thus, the co-ordinate of the particle is (4, 4).
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A cyclotron is a kind of particle accelerator invented by Ernest O. Lawrence at the University of California, Berkeley, in 1929–1930, and patented in 1932. A cyclotron accelerates charged particles along a spiral path outward from the center. The particles are guided by a static magnetic field to a spiral trajectory, and accelerated by a rapidly changing electric field (radio frequency). For this invention Lawrence received the 1939 Nobel Prize in Physics.Cyclotrons were the most efficient particle accelerator equipment until the 1950s, when the synchrotron replaced them, and are now used in science and nuclear medicine to generate particle beams.This will toss the particle out of resonance with the oscillating field which means that when the particle reaches the hole between the dees, the extremity of the dees won't change. Thus, the particle won't be accelerated further.So, Cyclotron isn't reasonable for accelerating electrons. Because of the small mass, the speed of electrons increments quickly. Similarly, because of fast relativistic variation in their mass, the electrons escape step with the oscillating electric field.Hence, option A is the correct option.

What will be the value of the co-ordinate whose position of a particle moving along the parabola y2 = 4x at which the rate at of increase of the abscissa is twice the rate of increase of the ordinate?a)(1, 1)b)(2, 2)c)(3, 3)d)(4, 4)Correct answer is option 'D'. Can you explain this answer?
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