Mathematics, 01.07.2019 07:00 Lesquirrel

Aparabolic microphone used on the sidelines of a professional football game uses a reflective dish 24 in. wide and 5 in. deep. how far from the bottom of the dish should the microphone b placed?

Aparabolic microphone used on the sidelines of a professional football game uses a reflective dish 2

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Answer from: liaholmes8

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(1 Question in written form): 7.2 inches.

(2 Question in the picture): Option (D) (y+1)^2=2(x+3) and $F=(\frac{-5}{2},-1)$ and $x=\frac{-7}{2}$

Explanations:

1. (Question in written form):

First you need to know the standard form of a parabola (considering it opens in +x direction and the vertex of the parabola is at origin), which is as follows:

(y-k)^2 = 4p(x-h)

Since the vertex of the parabola is at origin, h = 0 and k = 0. The above equation will now become:

y^2 = 4px ---- (A)

Where p is the distance between the vertex and the focus of a parabola, which we need to find (or where the microphone should be placed).

As the vertex is assumed to be placed at the origin, the parabolic dish, which is 24 inches wide, will be split evenly on both sides of the y axis (as the opening of the parabola is in x-axis—stated above), giving the distance of 12 inches from each sides of y axis (12 inches + 12 inches = 24 inches). Since it (parabolic microphone) is 5 inches deep, the coordinates will become: (5, +12) and (5, -12).

Plug any of the points—(5,+12) or (5,-12)—in equation (A), you will get the following:

$(12)^2 = 4p(5) \\ 144 = 4p(5) \\ \frac{144}{5} = 4p \\ \frac{144}{5*4} = p \\ p=7.2\thinspace inches$

Hence, the correct answer is 7.2 inches.

2. (Question in the picture):

First we need to write the following equation in the standard form:

y^2 - 2x + 2y -5 = 0 ----- (B)

The standard form of parabola is:

(y-k)^2 = 4p(x-h) ----- (C)

Rearrange equation (B) and solve as follows:

$y^2-2x+2y-5=0 \\ y^2 + 2y - 5 = 2x \\ y^2 + 2y +1 = 2x + 6 \\ y^2 + y + y + 1 = 2(x+3) \\ y(y+1)+1(y+1)=2(x+3)\\(y+1)^2=2(x+3)$

You can write the above equation as follows:

$(y-(-1))^2 = 4(\frac{1}{2})(x-(-3))$ ----- (D)

Now compare equation (D) with (C), we will get:

Focus = (h+p, k) = (-3+(1/2), -1)

$Focus: (\frac{-5}{2},-1)$

Now, to find x, use the following formula:

x = h - p

$x = -3-\frac{1}{2} = \frac{-7}{2}$

Hence, the correct option is (D) -> $F=(\frac{-5}{2},-1)$ and $x=\frac{-7}{2}$

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