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An arch is in the form of a parabola with its axis vertical. The arch is $10 \mathrm{~m}$ high and $5 \mathrm{~m}$ wide at the base. How side is it $2 \mathrm{~m}$ from the vertex of the parabola?
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$\because$ Axis of the parabola is vertical
$\therefore$ its equation is $x^2=4$ ay
Arch is $10 \mathrm{~m}$ high and $5 \mathrm{~m}$ wide at base
$\therefore$ Point $\left(\frac{5}{2}, 10\right)$ lies on parabola
$\therefore \quad \frac{25}{4}=4 \mathrm{a}(10)$
$\therefore \quad 4 a=\frac{5}{8}$
$\therefore$ equation of parabola becomes $x^2=\frac{5}{8} y$
Let width of arch $2 \mathrm{~m}$ from vertex is $2 b$ then, point $(b, 2)$ lies on parabola $b=\frac{\sqrt{5}}{2}$
$\therefore \quad$ width of arch is $2 b=2 \cdot \frac{\sqrt{5}}{2} \mathrm{~m}=\sqrt{5} \mathrm{~m}$ $=2.23 \mathrm{~m}$ (approx)
$\therefore$ its equation is $x^2=4$ ay
Arch is $10 \mathrm{~m}$ high and $5 \mathrm{~m}$ wide at base
$\therefore$ Point $\left(\frac{5}{2}, 10\right)$ lies on parabola
$\therefore \quad \frac{25}{4}=4 \mathrm{a}(10)$
$\therefore \quad 4 a=\frac{5}{8}$
$\therefore$ equation of parabola becomes $x^2=\frac{5}{8} y$
Let width of arch $2 \mathrm{~m}$ from vertex is $2 b$ then, point $(b, 2)$ lies on parabola $b=\frac{\sqrt{5}}{2}$
$\therefore \quad$ width of arch is $2 b=2 \cdot \frac{\sqrt{5}}{2} \mathrm{~m}=\sqrt{5} \mathrm{~m}$ $=2.23 \mathrm{~m}$ (approx)
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