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 $D\left({\displaystyle \frac{5}{2}},-10\right)$ lies on the parabola, it will satisfy the given equation of the parabola.
Substitute $x={\displaystyle \frac{5}{2}},y=-10$ in the equation of parabola $x^2=-4ay$ .
  ${\left({\displaystyle \frac{5}{2}}\right)}^2=-4a(-10)$
 ${\displaystyle \frac{25}{4}}=40a$
 $a={\displaystyle \frac{25}{4\times 40}}={\displaystyle \frac{5}{32}}$
So, the equation of the given parabola becomes,
 $x^2=-4\left({\displaystyle \frac{5}{32}}\right)y$
 $x^2=-{\displaystyle \frac{5}{8}}y$
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 $x^2=-{\displaystyle \frac{5}{8}}y$ .
Substitute $y=-2$ in this equation ,
 $x^2=-{\displaystyle \frac{5}{8}}(-2)$
 $x^2={\displaystyle \frac{5}{4}}$
Taking the square root on both sides of the equation
 $x=\pm \sqrt{{\displaystyle \frac{5}{4}}}$
 $x=\pm {\displaystyle \frac{\sqrt{5}}{2}}$
Since the width represents the length, we will not consider the negative value of $x$ , hence $$x={\displaystyle \frac{\sqrt{5}}{2}}$$.
Now, since the width is $2x$ m, so
$$2x=2\times {\displaystyle \frac{\sqrt{5}}{2}}=\sqrt{5}\approx 2.23$$
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.