Question

a flashlight has a parabolic reflector that reflects light waves from a bulb located at the focus. the figure shows a vertical cross section of such a parabolic reflector with the bulb at the focus. suppose the bulb is 0.2 in from the vertex of the reflector, and the distance across the widest part of the reflector is 4 in. suppose also that the parabola formed by the cross section is located on a grid so that (the figure may not be drawn to scale.) the vertex is at (0,0). the equation for the parabola is x =

Explanation:

Step1: Determine the standard form of the parabola

Since the parabola opens to the right and has its vertex at \((0,0)\), the standard form of the parabola is \(x = \frac{1}{4p}y^{2}\), where \(p\) is the distance from the vertex to the focus. Given that \(p = 0.2\).

Step2: Substitute the value of \(p\) into the equation

Substituting \(p = 0.2\) into \(x=\frac{1}{4p}y^{2}\), we get \(x=\frac{1}{4\times0.2}y^{2}\).

Step3: Simplify the coefficient

Calculate \(\frac{1}{4\times0.2}=\frac{1}{0.8}=\frac{10}{8}=\frac{5}{4}\).

Answer:

\(x = \frac{5}{4}y^{2}\)