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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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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?

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The problem can be diagrammatically represented as follows:
Since its a parabola opening upwards,its equation is $x^2 = 4ay$.
The parabola passes through the point $ (\large\frac{5}{2}$$,10)$
Therefore, $\large(\frac{5}{2})^2 $$=4\times a \times 10 \rightarrow a = \large\frac{5}{32}$
Therefore, the equation of the parabola is $x^2 = 4\times\large\frac{5}{32}$$\times y = \large\frac{5y}{8}$
We have to solve for $y=2m \rightarrow x^2 = \large\frac{5 \times 2}{8}$$ \rightarrow x = \large\sqrt {\frac{5}{4}}$
$\qquad AB = 2x = 2 \large\sqrt {\frac{5}{4}}$$ \approx 2.23m$
answered Apr 8, 2014 by balaji.thirumalai
 

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