# Thread: Lamp with a parabolic reflector

1. ## Lamp with a parabolic reflector

I attempted to solve this problem for hours.
I used x^2=4py and substitution. Didn't quite work.

A lamp with a parabolic reflector is shown in the figure. The bulb is placed at the focus and the focal diameter is 20 cm. Find the diameter of the opening 15 cm from the vertex. Round to the nearest centimeter, if necessary.
35
38
40
39
37

2. Originally Posted by jeunesse27
I attempted to solve this problem for hours.
I used x^2=4py and substitution. Didn't quite work.

A lamp with a parabolic reflector is shown in the figure. The bulb is placed at the focus and the focal diameter is 20 cm. Find the diameter of the opening 15 cm from the vertex. Round to the nearest centimeter, if necessary.
Focal diameter = 4p = 20 cm. So p = 5 cm
If F is the focus, FD = DM where M is the point on the diretrix. DM = 15 + 5 = 20. So FD = 20 cm.
Distance between the focus and center of opening circle is 15 - 5 = 10 cm.
So the radius R = 20^2 - 10^2 = 17.32. Hence
d = 2*17.32 = 34.63 ~ 35 cm.

3. Hello, jeunesse27!

Who wrote this problem?
The term "diameter" is usually reserved for circles.

A lamp with a parabolic reflector is shown in the figure. . Right!
The bulb is placed at the focus and the focal diameter is 20 cm. .
latus rectum?
Find the width $CD$ of the opening 15 cm from the vertex, to the nearest cm.

. . $(a)\;35 \qquad (b)\;38 \qquad (c)\;40 \qquad (d)\;39\qquad (e)\;37$
I'll take a guess at what's going on . . .
Code:
                |
15|
C o - - - - + - - - - o D
|
F|
(-10,p)o- - - - o - - - -o (10,p)
|
*       |p      *
*      |      *
*    |    *
- - - - -*o*- - - - - - - - - -
V

The equation is: . $x^2 \:=\:4py$
. . where $p$ is the distance from the vertex $V$ to the focus $F.$

The length of the latus rectum is $4p.$
. . Hence: . $p \,=\,5$

The equation is: . $x^2 \:=\:20y$

If $y = 15$, we have: . $x^2 \:=\:20(15) \:=\:300 \quad\Rightarrow\quad x \:=\:\pm\sqrt{300}\:=\:\pm10\sqrt{3}$

Therefore: . $CD \;=\;2(10\sqrt{3}) \;=\;34.64191615 \;\approx\;35\text{ cm}\;\;\;\text{ answer }(a)$

4. ## Thank you!

Originally Posted by sa-ri-ga-ma
Focal diameter = 4p = 20 cm. So p = 5 cm
If F is the focus, FD = DM where M is the point on the diretrix. DM = 15 + 5 = 20. So FD = 20 cm.
Distance between the focus and center of opening circle is 15 - 5 = 10 cm.
So the radius R = 20^2 - 10^2 = 17.32. Hence
d = 2*17.32 = 34.63 ~ 35 cm.
Thanks to all of my helpers! The detailed explanation cleared the topic up for me!

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# ind the diameter d(C, D) of the opening, 14 cm from the vertex. (Round your answer to the nearest whole number.)

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