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 x^{2} = 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, x^{2} = 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,

x^{2} = \( \dfrac{5}{8}(2)\)

= \( \dfrac{5}{4}\)

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 agoAn equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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