An arch is in the form of a parabola with its | vedclass

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(A) Let the vertex of the parabola be at the origin $(0, 0)$ and its axis be along the positive $y-$axis. The equation of the parabola is of the form

Vedclass · Accepted Answer

$2.23$

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(A) Let the vertex of the parabola be at the origin $(0, 0)$ and its axis be along the positive $y-$axis. The equation of the parabola is of the form $x^2 = 4ay$.The arch is $10 \, m$ high and $5 \, m$ wide at the base. This means the parabola passes through the point $(\frac{5}{2}, 10)$.Substituting this point into the equation:$(\frac{5}{2})^2 = 4a(10)$$\frac{25}{4} = 40a$$a = \frac{25}{160} = \frac{5}{32}$So,the equation of the parabola is $x^2 = 4(\frac{5}{32})y$,which simplifies to $x^2 = \frac{5}{8}y$.We need to find the width of the arch at $y = 2 \, m$ from the vertex.$x^2 = \frac{5}{8} 	imes 2 = \frac{5}{4}$$x = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \approx \frac{2.236}{2} = 1.118 \, m$.The total width of the arch is $2x = 2 	imes 1.118 = 2.236 \, m \approx 2.23 \, m$.

An arch is in the form of a parabola with its axis vertical. The arch is $10 \, m$ high and $5 \, m$ wide at the base. How wide is it $2 \, m$ from the vertex of the parabola (in $, m$)?

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