The reflecting dish of a parabolic microphone has a cross-section in the shape of a parabola. The microphone itself is placed on the focus of the parabola. If the parabola is 40 inches wide and 20 inches deep, how far from the vertex should the microphone be placed?
Let the equation for the parabola be
y =ax^2
We know that (20,20) is on the curve, so
a = 1/20 and our equation is
y = 1/20 x^2
Now recall that the parabola x^2 = 4py has its focus p units from its vertex. Our parabola is
x^2 = 20y, making p=5
so the focus is 5 units from the vertex.
To determine how far from the vertex the microphone should be placed in a parabolic microphone, we need to understand the basic properties of a parabola.
A parabola is a U-shape curve that has an axis of symmetry. The focus of the parabola is a special point along the axis of symmetry. In the case of the parabolic microphone, the microphone is placed on the focus.
In this problem, we are given the dimensions of the cross-section of the reflecting dish, which has the shape of a parabola. The width of the parabola is 40 inches, which means the distance between any two points on the parabola that lie on a line perpendicular to the axis of symmetry is 40 inches. The depth of the parabola is 20 inches, which is the distance from the vertex (the lowest point on the parabola) to the line perpendicular to the axis of symmetry.
Since we know the dimensions of the parabola, we can determine the equation of the parabola. The standard form equation of a parabola with a vertical axis is given by:
y = a(x-h)^2 + k
where (h, k) represents the vertex of the parabola.
In this case, since the parabola is symmetric along the y-axis and the vertex is at the origin (0, 0), the equation becomes:
y = ax^2
Next, we need to find the value of the constant "a" in the equation. We know that the distance between any two points on the parabola that lie on a line perpendicular to the axis of symmetry is 40 inches, which means the distance between two arbitrary points (x1, y1) and (x2, y2) on the parabola satisfies the equation:
(x2 - x1)^2 + (y2 - y1)^2 = 40^2
Using this equation, we can substitute the values (x1, y1) = (0, 0) (the vertex of the parabola) and (x2, y2) = (x, ax^2) (an arbitrary point on the parabola) to solve for "a":
(x - 0)^2 + (ax^2 - 0)^2 = 40^2
Simplifying this equation gives:
x^2 + a^2x^4 = 40^2
Since we know the depth of the parabola is 20 inches, we can substitute the value (x, y) = (20, 20) into the equation to solve for "a":
20^2 + a^2(20^4) = 40^2
Simplifying this equation gives:
400 + a^2(160000) = 1600
Dividing both sides of the equation by 160000 gives:
a^2 = (1600 - 400)/160000
a^2 = 1200/160000
a^2 = 0.0075
Taking the square root of both sides, we find:
a = √0.0075
Now that we have determined the value of "a", we can substitute it into the equation y = ax^2 and find the y-coordinate of the focus of the parabola (which is where the microphone should be placed). Since the focus lies on the axis of symmetry, the x-coordinate of the focus is zero.
Therefore, substituting x = 0 into the equation y = ax^2 gives:
y = 0.0075 * 0^2
y = 0
This means that the y-coordinate of the focus (where the microphone should be placed) is zero. Thus, the microphone should be placed at a distance of 0 inches from the vertex.
We can use the standard equation for a parabola with its vertex at the origin:
y = (1/4p) x^2
where p is the distance from the vertex to the focus. In this case, we know that the parabola is 40 inches wide, which means that the distance between the two points where the parabola intersects the x-axis (called the "latus rectum") is also 40 inches. That means:
4p = 40
p = 10
So the distance from the vertex (which is at the origin) to the focus is 10 inches. Since the microphone is placed at the focus, it should be 10 inches away from the vertex.