Question:

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?

Updated On: Oct 24, 2023
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Solution and Explanation

The origin of the coordinate plane is taken at the vertex of the arch in such a way that its vertical axis is along the positively-axis. 
This can be diagrammatically represented as

 arch in the form of a parabola with its axis vertical

The equation of the parabola is of the form \(x^ 2 = 4ay\)  (as it is opening upwards).
It can be clearly seen that the parabola passes through point \((\frac{5}{2}, 10)\)

\((\frac{5}{2})^2 = 4a(10)\)

\(4a = \frac{25}{(4\times4\times10)}\)

\(⇒ a =\frac{ 5}{32}\)

Therefore, the arch is in the form of a parabola whose equation is  \(x^2 = \frac{5}{8} (2)\)

When \(y = 2 m, x^2 = \frac{5}{8} \times(2)\)

\(⇒ x^2 = \frac{5}{4}\)

\(x = \sqrt{(\frac{5}{4})} = \sqrt{\frac{5}{2}}\)

\(∴ AB = 2 \times \sqrt{\frac{5}{2}}m = \sqrt{5}m = 2.23m\)(approx.)

Hence, when the arch is 2 m from the vertex of the parabola, its width is approximately 2.23 m.

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Concepts Used:

Parabola

Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).

Parabola


 

 

 

 

 

 

 

 

 

Standard Equation of a Parabola

For horizontal parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A,
  1. Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
  2. P(x,y) as the moving point.
  • Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
  • The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
  • The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
  • By definition PM = PS

=> MP2 = PS2 

  • So, (a + x)2 = (x - a)2 + y2.
  • Hence, we can get the equation of horizontal parabola as y2 = 4ax.

For vertical parabola

  • Let us consider
  • Origin (0,0) as the parabola's vertex A
  1. Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
  2. P(x,y) as any moving point
  • Let us now draw a perpendicular SZ from S to the directrix.
  • Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
  • The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
  • By definition PM = PS

=> MP2 = PS2

So, (b + y)2 = (y - b)2 + x2

  • As a result, the vertical parabola equation is x2= 4by.