a manufacturer is designing a flashlight for the flashlight to admit a focused beam the bull needs to be on the central axis of the parabolic reflector 33 cm from the vertex write an equation that models the parabola formed when a cross section is taken through the reflector Center access assume that the vertex of the parabola is at the origin of the XY coordinate plane in the parabola can open it any direction

The formula for a veritcal parabola is y=1/4p(x-h)^2+k
(h,k)= coordinates of the vertex of the parabola
p= absolute value of the distance from the vertex to the focus/directrix
In this problem, it is given that the vertex is at the origin (0,0) and the focus (the bulb), is 3 centimeters away from the vertex.
Now, you know the values of the variables. Fill in the values
FROM THE FORMULA: 1/4p turns into 1/12 since p is 3.
(x-h)^2+k turns into (x-0)^2+0, since h and k where the values of the vertex which was 0,0
once all the variables are given values (except x and y) you have made your equation!
The answer is y=1/12(x-0)^2+0

MARK BRAINLIEST! this took forever, and I know this is the right answer since I got a perfect grade when I turned mine in.

oeerivona oeerivona · Answer 2 · 2021-08-15T15:06:47+02:00

The equation of a parabola and the coordinates of the focus can be

expressed in terms of the coordinate of the vertex and the coefficient of

the quadratic term

The equation that models the parabola formed by taking a cross section

through the Center of the flashlight with focus 33 cm from the vertex is [tex]y \ \mathbf{ = \dfrac{x^2}{132} }[/tex]

The process by which the above equation is arrived at is as follows;

The given parameters of the parabolic reflector are;

The focus of the parabolic cross section of the reflector = 33 cm from the

vertex, along the central axis

The location of the vertex of the parabolic cross section = The origin of the XY coordinate plane = (0, 0)

The vertex form of the equation of a parabola is y = a·(x - h)² + k

Where;

(h, k) = The coordinate of the vertex of the parabola

The coordinate of the focus, f = (h, k + 1/(4·a))

In the given parabola, the vertex (h, k) = (0, 0)

Therefore;

h = 0, and k = 0

The vertex of a parabola is along the central axis, therefore, the equation

of the central axis is the vertical line, x = 0

The x-coordinate of the focus, is therefore, x = 0

From which from the distance of the focus from the vertex = The y-coordinate = 33

The coordinate of the focus is (0, 33)

Equating the above coordinate to the general form of the coordinate of a parabola (opening up or down), gives;

Coordinate of the focus, f = (h, k + 1/(4·a)) = (0, 33)

Therefore;

As above, h = 0

k + 1/(4·a) = 33

From k = 0 obtained above, we get;

0 + 1/(4·a) = 33

a = 1/(4 × 33) = 1/132

a = 1/132

The equation of the parabola is therefore found by plugging in the values

of a, h, and k in the equation, y = a·(x - h)² + k, as follows;

[tex]y = \dfrac{1}{132} \times \left(x - 0\right )^2 + 0 \ \mathbf{ = \dfrac{x^2}{132} }[/tex]

[tex]\therefore \mathbf{The \ equation \ of \ the \ parabola \ is \ y} \mathbf{ = \dfrac{x^2}{132} }[/tex]

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