1 Answer
Solution :-
We know that the origin of the coordinate plane is taken at the vertex of the arch, where its vertical axis is along the positive y –axis.
Diagrammatic representation is as follows:
The equation of the parabola is of the form x2 = 4ay (as it is opening upwards).
It is given that at base arch is 10m high and 5m wide.
So, y = 10 and x = \( \dfrac{5}{2}\) from the above figure.
It is clear that the parabola passes through point (\( \dfrac{5}{2}\), 10)
So, x2 = 4ay
\( (\dfrac{5}{2})^2\) = 4a(10)
4a = \( \dfrac{25}{(4×10)}\)
a = \( \dfrac{5}{32}\)
we know the arch is in the form of a parabola whose equation is
\( x^2=\dfrac{5}{8}y\)
We need to find width, when height = 2m.
To find x, when y = 2.
When, y = 2,
x2 = \( \dfrac{5}{8}(2)\)
= \( \dfrac{5}{4}\)
x = \(\sqrt{\dfrac{5}{4}}\)
= \(\sqrt{\dfrac{5}{2}}\)
AB = 2 × \(\sqrt{\dfrac{5}{2}}\)m
= \( \sqrt{5}m\)
= 2.23m (approx.)
Hence, when the arch is 2m from the vertex of the parabola, its width is approximately 2.23m.
Answered by Abhisek | 1 year agoRelated Questions
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