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(A) Let the vertex of the parabola be at the lowest point $O(0,0)$ and the axis be vertical along the $Y$-axis. The equation of the parabola is $x^2 =

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$2 \sqrt{6} \, m$

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(A) Let the vertex of the parabola be at the lowest point $O(0,0)$ and the axis be vertical along the $Y$-axis. The equation of the parabola is $x^2 = 4ay$.The total span is $12 \, m$,so the ends are at $x = 6 \, m$ and $x = -6 \, m$. The maximum deflection at the centre is $3 \, cm = 0.03 \, m$. Thus,the parabola passes through $(6, 0.03)$.Substituting these coordinates into the equation: $(6)^2 = 4a(0.03) \implies 36 = 0.12a \implies a = \frac{36}{0.12} = 300$.The equation is $x^2 = 4(300)y = 1200y$.We want to find $x$ when the deflection from the top is $1 \, cm$. Since the total deflection is $3 \, cm$,the height from the vertex $O$ is $y = 3 \, cm - 1 \, cm = 2 \, cm = 0.02 \, m$.Substituting $y = 0.02$ into the equation: $x^2 = 1200(0.02) = 24$.Therefore,$x = \sqrt{24} = 2\sqrt{6} \, m$.

$A$ beam is supported at its ends by supports which are $12 \, m$ apart. Since the load is concentrated at its centre,there is a deflection of $3 \, cm$ at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1 \, cm$?

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