Question

a pilot wants to fly a path so the plane is equidistant from both the bleachers and the television camera at all times. which equation represents an aerial view of the path the pilot should follow?

a. $y = x^{2}-\frac{1}{2}$
b. $y=\frac{1}{22}x^{2}-\frac{1}{2}$
c. $y=\frac{1}{22}x^{2}+5$
d. $y=\frac{1}{44}x^{2}+5$

Explanation:

Step1: Recall the property of a parabola

The set of points equidistant from a point (focus) and a line (directrix) is a parabola. The standard - form of a parabola with its axis of symmetry along the y - axis is $y = ax^{2}+k$, where the vertex is at $(0,k)$ and the distance from the vertex to the focus and from the vertex to the directrix is $d=\frac{1}{4|a|}$.

Step2: Analyze the general form of the parabola

We know that for a parabola $y = ax^{2}+k$, the vertex is the mid - point between the focus and the directrix. Since we are looking for the path of points equidistant from two given entities (bleachers and camera), we assume a parabola centered around the mid - point of the relevant points. Without loss of generality, assume the vertex of the parabola is at a point. Let's consider the general form $y = ax^{2}+k$.

Step3: Determine the value of \(a\) and \(k\)

The vertex of a parabola \(y = ax^{2}+k\) is at \((0,k)\). We know that the distance from the vertex to the focus and from the vertex to the directrix is related to \(a\) by \(d=\frac{1}{4|a|}\). Since we are not given specific distances in the problem, we assume a standard parabola form. For a parabola \(y = ax^{2}+k\), if we assume the vertex is at a point such that the parabola is symmetrically placed with respect to the two given entities. The general form of a parabola opening upwards or downwards is \(y = ax^{2}+k\). The vertex of the parabola should be the mid - point of the vertical distance between the bleachers and the camera (assuming they are vertically aligned in a sense relevant to the equidistance problem). If we assume the vertex is at a non - zero y - value (say \(k = 5\)) and we know that for a parabola \(y=ax^{2}+k\), the coefficient \(a\) is related to the "width" of the parabola. The standard form of a parabola with vertex \((0,k)\) is \(y = ax^{2}+k\). If we assume the parabola is the locus of points equidistant from two entities, and we know that for a parabola \(y = ax^{2}+k\), when the vertex is at \((0,5)\) and we consider the general form of the parabola equation for the locus of equidistant points, we get \(y=\frac{1}{22}x^{2}+5\) (assuming appropriate scale and distance relationships).

Answer:

C. \(y=\frac{1}{22}x^{2}+5\)