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A searchlight uses a parabolic mirror to cast a beam of light in parallel rays where the light bulb is located at the focus of the parabola in order to give the best illumination. The mirror is modeled by y2 = 20(x – 10), where the measurements are in cm. What is the location of the light bulb?



Answer :

jacob193

Answer:

[tex](15,\, 0)[/tex].

Step-by-step explanation:

In general, depending on whether the parabola opens vertically or horizontally, the the vertex equation of a parabola would be:

  • [tex](x - h)^{2} = 4\, p\, (y - k)[/tex] if the parabola opens vertically, or
  • [tex](y - k)^{2} = 4\, p\, (x - h)[/tex], if the parabola opens horizontally.

In both equations, [tex]k[/tex], [tex]h[/tex], and [tex]p[/tex] are constants, where [tex]h[/tex] and [tex]k[/tex] specify the position of the vertex of the parabola, and [tex]p[/tex] specifies the distance between the vertex and the focus of the parabola.

While the vertex of the parabola is at [tex](h,\, k)[/tex] regardless of whether the parabola opens vertically or horizontally, the position of the focus of the parabola depends on the direction of the opening:

  • If the parabola opens vertically, the focus would be either above (if [tex]p > 0[/tex]) or below (if [tex]p < 0[/tex]) the vertex, at [tex](h,\, k + p)[/tex].
  • If the parabola opens horizontally, the focus would be either to the left (if [tex]p < 0[/tex]) or to the right (if [tex]p > 0[/tex]) the vertex, at [tex](h + p,\, k)[/tex].

Notice that in the two equations, only one of [tex]x[/tex] or [tex]y[/tex] is raised to the power of [tex]2[/tex], but not both. Whether the parabola opens vertically or horizontally depends on the variable that is raised to a power of [tex]2[/tex]: vertically if [tex]x[/tex] is raised to a power of [tex]2[/tex], and horizontally if [tex]y[/tex] is raised to a power of [tex]2[/tex].

In the parabola in this question, [tex]y[/tex] is raised to a power of [tex]2[/tex]. Hence, this parabola would open horizontally. Rearrange the given equation to match the form of the vertex equation for a parabola that opens horizontally:

[tex]y^{2} = 2\, (x - 10)[/tex].

[tex](y - 0)^{2} = 4\, (5)\, (x - 10)[/tex].

Hence, [tex]k = 0[/tex], [tex]h = 10[/tex], and [tex]p = 5[/tex] for this parabola.

Since this parabola opens horizontally and [tex]p > 0[/tex], the focus would be located to the right of the vertex, at [tex](h + p,\, k)[/tex], which would be [tex](15,\, 0)[/tex].

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