Question

2. A flashlight is shaped like a paraboloid, so that if its light bulb is placed at
the focus, the light rays from the bulb will then bounce of the surface in a
focused direction that is parallel to the axis. If the paraboloid has a depth of
1.8 in and the diameter on its surface is 6 in, how far should the light source
be placed from the vertex?​

Answer 1

Answer:

Brainiest me

Step-by-step explanation:

The light bulb should be placed at a distance of 1 inch from the paraboloid vertex.

Answer 2

$(0,-1.8+1.25). (0,-0.55)$

Above values are the light source be placed from the vertex.

Find distance between light source and vertex :

Let's put the paraboloid's area on the x-axis as well as the paraboloid's center on the y-axis.

The vertex is at 1.8 inches if the depth is 1.8 inches (0,-1.8)

The equation for a vertically opening parabola is as follows:

• $(x-h)2 = 4p(y-k)$, where the vertex is (h,k).

• To find p, we can choose one of the two locations near the edge of the surface [(-3,0) and (3,0).

• $x^2 = 4p(y+1.8) - > 32$
• $4p(1.8) - > 9 = 7.2p - > p = 1.25$

• The parabola's center is situated at:

• Because F = (h,k+p), the light source should be located at $(0,-1.8+1.25). (0,-0.55)$