Parabolic Headlight where should bulb be located Locus and Conic unit

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SUMMARY

The discussion centers on the placement of a bulb in a parabolic headlight reflector, specifically at the focus of the parabola. The equation provided, \(y^2 = 4px\), indicates that the bulb should be positioned at the point (8, 0), where 8 represents the focus of the parabola. Participants clarify that the parabola opens to the right, and the dimensions of the reflector confirm this placement. The correct interpretation of the parabola's geometry is crucial for optimal light reflection.

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  • Understanding of parabolic equations, specifically \(y^2 = 4px\)
  • Familiarity with the concept of focus in conic sections
  • Basic knowledge of coordinate geometry
  • Ability to interpret diagrams related to conic sections
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  • Study the properties of parabolas and their applications in optics
  • Learn how to derive the focus of a parabola from its equation
  • Explore practical applications of parabolic reflectors in lighting design
  • Investigate the relationship between the dimensions of a parabola and its focal point
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http://ca.pg.photos.yahoo.com/ph/sikandar1984/detail?.dir=9181&.dnm=8e2e.jpg&.src=ph

This link is a diagram for this question

It says the cross section of a parabolic reflector of a headlight is shown in the diagram. Determine where the bulb should be located if it is positioned at the focus?

Um what is the cross section? The x and y axis? Where they meet?

Do I have to find an equation? Or coordinate? Or do I just describe where it should be?

This is an application question for locus and conics I think this is an application with a parabola conic but not too sure. I have an equation for the bulb but its probably wrong, it is

x^2= \frac {1} {32} y^2

this is the equation of a parabola conic opening towards the right like the diagram. Please help me out! Thanks
 
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aisha said:
x^2= \frac {1} {32} y^2

this is the equation of a parabola conic opening towards the right like the diagram. Please help me out! Thanks
The equation for this parabola is:

y^2 = 4px where p is the focus

So if the point (8,16) is on the parabola, what is p?

That is where you would put the light bulb. At that point all lines to the parabola reflect parallel to the x-axis - which makes it a beam of light.

[Note: the parabola is obviously not drawn to the proper scale.]

AM
 
Last edited:
where did u get the point (8,16) is this parabola opening towards the right like the diagram?
 
aisha said:
where did u get the point (8,16) is this parabola opening towards the right like the diagram?

Your picture shows that at 8 units from the vertex, the "width" of the parabola is 32: If your coordinate system is set up so that (0,0) is at the vertex then 1/2 of that 32 is positive, 1/2 negative. Choosing positive x to the right, positive y upward, that corresponds to x= 8, y= 32/2= 16 or y= -32/2= -16.

aisha said:
This is an application question for locus and conics I think this is an application with a parabola conic but not too sure.

aisha, Don't you believe what YOU write? The problem SAYS "a parabolic reflector". Why would you be "not too sure" it really IS a parabola?
 
the light bulb will be at point 8,0

8 being the focus when i subbed a point into the equation :smile:
 

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