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=\frac{5}{4}y^{2} ). the depth of the reflector is ( square ) in.

Explanation:

Step1: Determine the value of \( y \)

The distance across the widest part of the reflector is \( 4 \) in. Since the parabola is symmetric about the \( x - \)axis, if the total width is \( 4 \) in, then \( y = 2 \) (taking the positive value for the upper - half of the parabola).

Step2: Substitute \( y = 2 \) into the parabola equation

The equation of the parabola is \( x=\frac{5}{4}y^{2}\).
Substitute \( y = 2 \) into the equation:

$$
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$$

Answer:

\( 5 \)