1. **State the problem:** We have the parabola given by the equation $$(y+4)^2 = \frac{1}{2}(x-2)$$ and we want to find the equation of this parabola after rotating it by an angle of 60° anticlockwise. 2. **Recall the rotation formulas:** When we rotate a point $(x,y)$ about the origin by an angle $\theta$ anticlockwise, the new coordinates $(x',y')$ are given by: $$x' = x\cos\theta - y\sin\theta$$ $$y' = x\sin\theta + y\cos\theta$$ 3. **Apply rotation to the parabola:** Here, $\theta = 60^\circ = \frac{\pi}{3}$ radians. We have: $$\cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2}$$ 4. **Express original variables in terms of rotated variables:** From the rotation formulas, solving for $x$ and $y$: $$x = x'\cos 60^\circ + y'\sin 60^\circ = \frac{1}{2}x' + \frac{\sqrt{3}}{2}y'$$ $$y = -x'\sin 60^\circ + y'\cos 60^\circ = -\frac{\sqrt{3}}{2}x' + \frac{1}{2}y'$$ 5. **Substitute into the original parabola equation:** Original equation: $$(y+4)^2 = \frac{1}{2}(x-2)$$ Substitute $x$ and $y$: $$\left(-\frac{\sqrt{3}}{2}x' + \frac{1}{2}y' + 4\right)^2 = \frac{1}{2}\left(\frac{1}{2}x' + \frac{\sqrt{3}}{2}y' - 2\right)$$ 6. **Simplify the right side:** $$\frac{1}{2}\left(\frac{1}{2}x' + \frac{\sqrt{3}}{2}y' - 2\right) = \frac{1}{4}x' + \frac{\sqrt{3}}{4}y' - 1$$ 7. **Final rotated parabola equation:** $$\left(-\frac{\sqrt{3}}{2}x' + \frac{1}{2}y' + 4\right)^2 = \frac{1}{4}x' + \frac{\sqrt{3}}{4}y' - 1$$ This is the equation of the parabola after rotation by 60° anticlockwise. **Note:** The variables $x', y'$ represent the coordinates after rotation.