Solve the problem.
-A domed ceiling is a parabolic surface. For the best lighting on the floor, a light source is to be placed at the focus of the surface. If $5 \mathrm {~m}$ down from the top of the dome the ceiling is $23 \mathrm {~m}$ wide, find the best location for the light source.
A) $26.5 \mathrm {~m}$ down from the top
B) $19.8 \mathrm {~m}$ down from the top
C) $13.2 \mathrm {~m}$ down from the top
D) $6.6 \mathrm {~m}$ down from the top

Question

Quizplus · Accepted Answer

The equation of a parabola in vertex form is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex of the parabola. For a parabolic dome, the vertex is at the top of the dome. If we place the origin of our coordinate system at the vertex of the dome, the equation simplifies to $y = ax^2$, since $h = 0$ and $k = 0$. The focus of a parabola is located at $(0, \frac{1}{4a})$, where $a$ is the coefficient in the parabola's equation.Given that the dome is $23 \, \text{m}$ wide when $y = 5 \, \text{m}$, we can use the information to find $a$. The width of $23 \, \text{m}$ means that from the center of the dome (vertex), the parabola extends $11.5 \, \text{m}$ to the left and right. Therefore, when $x = 11.5$, $y = 5$. Substituting these values into the equation $y = ax^2$, we get:$$5 = a(11.5)^2$$Solving for $a$, we find:$$a = \frac{5}{(11.5)^2} = \frac{5}{132.25}$$The focus of the parabola is located at $(0, \frac{1}{4a})$, so we calculate:$$\frac{1}{4a} = \frac{1}{4 \cdot \frac{5}{132.25}} = \frac{132.25}{20} = 6.6125 \, \text{m}$$Therefore, the best location for the light source is $6.6 \, \text{m}$ down from the top of the dome, which corresponds to choice D.

Solve the Problem 5 m5 \mathrm {~m}5 m Down from the Top of the Dome the Ceiling Is

Solve the Problem $5 \mathrm {~m}$ Down from the Top of the Dome the Ceiling Is