Question

A particle is moving with a velocity ⃗v = K(yî + xjˆ)
, where K is a constant. The general equation for its path is:
(A) y = x² + constant (B) y² = x + constant
(C) y² = x² + constant (D) xy = constant
[JEE Main 2019]

Answer 1

Answer:

C. will be correct answers

Answer 2

The general equation for the path of the particle is

• $y^{2} = x^{2} +$constant

, its given that

• V = k(yi + xj), velocity is rate of change of position

• Therefore, Vx = ky

•                  Vy = kx

•        We know, Vx = $\frac{dx}{dt}$  

•     and            Vy = $\frac{dy}{dt}$  

Applying this above, we get:

• $\frac{dx}{dt} = ky$

• $\frac{dy}{dt} = kx$
• dx/ky = dt
• dy/kx = dt ==>
• dt = dx/ky = dy/kx ==>
• Rearranging both sides to integrate:
• kxdx = kydy ==>
• xdx = ydy ==.>
• Integrating both sides ==>
• $x^{2} = y^{2}$ + constant or $y^{2} = x^{2} +$constant
• Answer is option C