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.

Answer 1

Hint: A parabola is defined as a set of points that are equidistant from a directrix, which is a fixed straight line and the focus. If the parabola has directrix as the x-axis, and the focus is $ (a,0) $ , then the equation of the parabola is given by $ {y^2} = 4ax $ and if the parabola has directrix as the y-axis, and the focus is $ (0,a) $ , then the equation of the parabola is given by $ {x^2} = 4ay $ . If any point lies on the parabola, it means that it will satisfy the equation of the given parabola.

Complete step by step answer:
It is given that an arch is in the form of a parabola with its axis vertical and the arch is $ 10 $ m high and $ 5 $ m wide at the base. So, to illustrate it in the form of a figure, let us take the vertex of this parabola to be at origin $ (0,0) $ . Then it will form a parabola such that its vertical is at origin and the directrix is along the negative y-axis. To represent it diagrammatically, we have,

From the given figure, we see that the equation of the parabola opening on the negative y axis is given by $ {x^2} = - 4ay $ .
We need to determine the value of the focus, that is $ a $ for the given parabola, to proceed further.
Since the point $ D\left( {\dfrac{5}{2}, - 10} \right) $ lies on the parabola, it will satisfy the given equation of the parabola.
Substitute $ x = \dfrac{5}{2},y = - 10 $ in the equation of parabola $ {x^2} = - 4ay $ .
  $ {\left( {\dfrac{5}{2}} \right)^2} = - 4a( - 10) $
 $ \dfrac{{25}}{4} = 40a $
 $ a = \dfrac{{25}}{{4 \times 40}} = \dfrac{5}{{32}} $
So, the equation of the given parabola becomes,
 $ {x^2} = - 4\left( {\dfrac{5}{{32}}} \right)y $
 $ {x^2} = - \dfrac{5}{8}y $
Now to determine the width of the arch, when measured $ 2 $ m from the vertex of the parabola, say the width is $ 2x $ m. From the given parabola figure, we need to determine the value of $ x $ , when measured $ 2 $ m away from the x-axis. That is, we are required to determine the coordinates of the point $ B(x, - 2) $ . Here, $ - 2 $ represent that the point $ B $ is towards the negative side of the y-axis, hence $ y = - 2 $ .
Since this point $ B $ lies on the parabola, it will satisfy the equation of the given parabola represented by $ {x^2} = - \dfrac{5}{8}y $ .
Substitute $ y = - 2 $ in this equation ,
 $ {x^2} = - \dfrac{5}{8}( - 2) $
 $ {x^2} = \dfrac{5}{4} $
Taking the square root on both sides of the equation
 $ x = \pm \sqrt {\dfrac{5}{4}} $
 $ x = \pm \dfrac{{\sqrt 5 }}{2} $
Since the width represents the length, we will not consider the negative value of $ x $ , hence \[x = \dfrac{{\sqrt 5 }}{2}\].
Now, since the width is $ 2x $ m, so
\[2x = 2 \times \dfrac{{\sqrt 5 }}{2} = \sqrt 5 \approx 2.23\]
So, the width of the arc is approximately $ 2.23 $ when measured $ 2 $ m from the vertex of the parabola.

Note:
For a parabola having the equation $ {x^2} = 4ay $ , the axis of symmetry is the y-axis and vertex lie on the origin. And for a parabola having the equation $ {y^2} = 4ax $ , the axis of symmetry is the x-axis and vertex lie on the origin. If the value of $ a $ is positive, then the parabola will have focus on the positive side of the axis of symmetry, and if the value of $ a $ is negative, then the parabola will have focus on the negative side of the axis of symmetry.