$ x = 2 \sin \omega t \,;$ $ y = 2 \sin \left( {\omega t + \frac{\pi }{4}} \right)$
The path of the particle will be :
$\cos \omega t=\frac{\sqrt{4-x^{2}}}{2}$
$y=2 \sin \omega t+\frac{2}{\sqrt{2}} \cos \omega$ ltbgt $=\sqrt{2} \times \frac{x}{2}+\sqrt{2} \frac{\sqrt{4-x^{2}}}{2}$
$=\frac{x}{\sqrt{2}}+\left(\frac{\sqrt{4-x^{2}}}{\sqrt{2}}\right)$
or $\sqrt{2} y-x=\sqrt{4-x^{2}}$
or $2 y^{2}+x^{2}-x y=4-x^{2}$
or $2 y^{2}+2 x^{2}-x y=4$
or $x^{2}+y^{2}-\sqrt{2} x y=2$
This is an equation of ellipse.
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