A parabolic headlight is formed by revolving the parabola y^2 = 15x between the lines y = -3 and y = 3 about its line of symmetry. Where should the headlight bulb be placed for maximum illumination?

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Maximum illumination will be achieved if the bulb is placed at the focus of the parabola created by the cross-section of the headlight.

If we rearrange the equation we know that:

`x=1/15y^2` , therefore, a = 1/15

The focus of a parabola is located at:

`(h,k+1/(4a))`

If we place the vertex of the parabola at the origin (0,0) for simplicity this gives:

`(0,1/(4a))`

Substituting our known value of a we get:

`(0,1/(4(1/15)))=(0,3.75)`

Therefore, the bulb should be placed 3.75 units to the right of the vertex.

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