Q.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?

The key idea is to model the arch as a downward-opening parabola with vertex at the origin, using the standard form $x^2 = -4ay$.

Step 1: Place the vertex at $(0,0)$ and the axis along the $y$-axis. The parabola opens downward, so its equation is $x^2 = -4ay$.

Step 2: The base is $5$ m wide and $10$ m below the vertex. Thus the point $(2.5, -10)$ lies on the parabola. Substitute:

$(2.5)^2 = -4a(-10) \implies 6.25 = 40a \implies a = \frac{6.25}{40} = 0.15625$.

Step 3: The equation becomes $x^2 = -4(0.15625)y = -0.625y$. We need the width $2$ m from the vertex, i.e., at $y = -2$.

Substitute: $x^2 = -0.625(-2) = 1.25 \implies x = \sqrt{1.25} = \frac{\sqrt{5}}{2}$.

The total width at that height is $2x = \sqrt{5}$ m.

The problem gives us an arch shaped like a parabola with its axis vertical. That means the parabola opens either upward or downward. Since an arch is highest at the middle and curves down to the base, the vertex is at the top and the parabola opens downward. This is the key insight: the vertex is the highest point, and the base is symmetric about the vertical axis.

We place the vertex at the origin $(0,0)$ for convenience. The axis is the $y$-axis, so the parabola opens downward along the negative $y$-direction. The standard form for such a parabola is:

Here $a > 0$ is the distance from the vertex to the focus. The negative sign tells us the parabola opens downward.

Now we use the given dimensions to find $a$.

1. Interpret the given data.

   The arch is 10 m high — that means from the vertex (top) to the base (bottom) the vertical distance is 10 m. So the base lies on the horizontal line $y = -10$.

   The arch is 5 m wide at the base — that means the two endpoints of the base are 5 m apart horizontally. Since the parabola is symmetric about the $y$-axis, each endpoint is $2.5$ m from the axis. So the base endpoints are $(\pm 2.5, -10)$.

2. Plug a base point into the equation.

   Take the point $(2.5, -10)$. It must satisfy $x^2 = -4ay$:

So the parabola equation is:

3. Find the width 2 m from the vertex. "2 m from the vertex" means the vertical distance from the vertex is 2 m. Since the vertex is at $y=0$ and the arch goes downward, this point is at $y = -2$. Substitute $y = -2$ into the equation:

So $x = \pm \frac{\sqrt{5}}{2}$.

The total width at this height is the distance between the two symmetric points: