Question

A parabolic microphone has a cross-section in the shape of a parabola. The microphone is placed at the focus of the parabola. If the parabola is 20 inches wide and 5 inches deep, how far from the vertex should the microphone be placed?

Answer 1

The equation of a parabola is
\[y ^{2}=4ax\]
where the focus is (a, 0) and the vertex is (0, 0)
In this problem the value of y is half the width of the parabola, and x = 5 therefore
\[(\frac{20}{2})^{2}=4\times a \times 5\ .........(1)\]
All you need to do is to solve equation (1) to find the value of a.

Answer 2

wait im still kind of confused a is what im solving?

Answer 3

Yes. Remember the focus is at the point (a, 0). The vertex is at point (0, 0), so a is the distance from the vertex where the microphone should be placed.

Answer 4

\[a=\frac{(\frac{20}{2})^{2}}{4\times 5}\]

Answer 5

i got 5 is that right?

Answer 6

Yes, your answer is right. Well done :)
So the microphone should be placed on the axis of the parabola spaced 5 inches from the vertex. The distance happens to be the same as the height of the parabola.

Answer 7

oh okay thank you!! do you think u could help me with some more practice problems?

Answer 8

Sure, I can help with one more before I must log out.

Answer 9

okay ill post it now

Answer 10

A hyperbolic mirror can be used to take panoramic photos, if the camera is pointed toward the mirror with the lens at one focus of the hyperbola. Write the equation of the hyperbola that can be used to model a mirror that has a vertex 4 inches from the center of the hyperbola and a focus 1 inch in front of the surface of the mirror. Assume the mirror has a horizontal transverse axis and the hyperbola is centered at (0, 0).

Answer 11

wait you therE?

Answer 12

I am looking into the question now. Please wait a little.

Answer 13

ok thank you

Answer 14

The general equation of the hyperbola is
\[\frac{x ^{2}}{a ^{2}}-\frac{y ^{2}}{b ^{2}}=1\]
The vertex is at a and a =4, therefore the equation of the model now becomes
\[\frac{x ^{2}}{16}-\frac{y ^{2}}{b ^{2}}=1\]
Now we need to find b. The foci of the hyperbola have coordinates at
\[(\pm \sqrt{a ^{2}+b ^{2}}, 0)\]
We know the position of the focus is 1 inch in front of the vertex, therefore the x coordinate of the focus = 4 + 1 = 5
So we can make an equation for the x coordinate of the focus as follows:
\[\sqrt{4^{2}+b ^{2}}=5^{2}\ ...................(1)\]
Can you solve (1) to find the value of b?

Answer 15

Sorry, my bad. Equation (1) should read
\[\sqrt{4^{2}+b ^{2}}=5\ ..............(1)\]
Can you solve (1) to find the value of b?

Answer 16

umm im kind of confused on how i would find it

Answer 17

Square both sides of the equation (1). What do you get?

Answer 18

sq it by 1?

Answer 19

Not really. When you square this what do you get:
\[\sqrt{9}\]

Answer 20

3

Answer 21

Well that is the square root of 9 sure enough. What I meant to ask is
\[(\sqrt{9})^{2}=\sqrt{9}\times \sqrt{9}=?\]

Answer 22

As you correctly wrote before
\[\sqrt{9}=3\]
So we get
\[(\sqrt{9})^{2}=\sqrt{9}\times \sqrt{9}=3\times 3=?\]

Answer 23

27?

Answer 24

Not so.
\[3\times 3=?\]

Answer 25

9 lol sry

Answer 26

What I am trying to show is the rule
\[(\sqrt{a})^{2}=a\] So getting back to equation (1)
\[(\sqrt{4^{2}+b ^{2}})^{2}=4^{2}+b ^{2}\]

Answer 27

So the result of squaring both sides of equation (1) is
\[4^{2}+b ^{2}=5^{2}.................(2)\]
Can you solve equation (2) to find the value of b?

Answer 28

yeah i can use pyth therom right?

Answer 29

Yes, you certainly can solve it that way.

Answer 30

so then is it 3?

Answer 31

Good work. You are correct. The constant b = 3.
Finally the equation to model the parabolis mirror becomes:
\[\frac{x ^{2}}{16}-\frac{y ^{2}}{9}=1\]

Answer 32

okay so then i am done?

Answer 33

Yes, you have solved the question :)

Answer 34

thank you soooo much :)!!

Answer 35

You're welcome :)

Answer 36

will you be on tomorrow so i can get some more help on my practice problems?

Answer 37

I expect that I will :)

Answer 38

hyperbolic mirror*