A particle is moving with a velocity $\vec{v}=K(y\hat{i}+x\hat{j})$, where K is a constant. The general equation for its path is:

(1) xy = constant

(2) y^{2} = x^{2} + constant

(3) y = x^{2} + constant

(4) y^{2} = x + constant

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Ans: (2) y^{2} = x^{2} + constant

Sol: ${dx\over dt}=ky, {dy\over dt}=kx$

Now, ${dy\over dx}={{dy\over dt}\over{dx\over dt}}={x\over y}$

$\implies ydy=xdx$

Integrating both side

$y^2=x^2+c$

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