MATH SOLVE

3 months ago

Q:
# Gabriella is designing a flashlight that uses a parabolic reflecting mirror and a light source. The shape of the mirror can be modeled by (y+3)^2=26(x-2) where x and y are measured in inches. Where should she place the bulb to ensure a perfect beam of light?

Accepted Solution

A:

the bulb should be placed at the "focus point" of the parabolic mirror.

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf (y+3)^2=26(x-2)\implies [y-\stackrel{k}{(-3)}]^2=\stackrel{4p}{26}(x-\stackrel{h}{2}) \\\\\\ 4p=26\implies p=\cfrac{26}{4}\implies p=\cfrac{13}{2}\\\\ -------------------------------\\\\ vertex~(2,-3)\qquad focus~\left(2+\frac{13}{2}~,~-3\right)[/tex]

check the picture below.

bear in mind that, because the leading term's coefficient is positive, namely for y², the parabola opens to the right, and "p" is positive, therefore the focus point will be to the right of the vertex.

[tex]\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------\\\\[/tex]

[tex]\bf (y+3)^2=26(x-2)\implies [y-\stackrel{k}{(-3)}]^2=\stackrel{4p}{26}(x-\stackrel{h}{2}) \\\\\\ 4p=26\implies p=\cfrac{26}{4}\implies p=\cfrac{13}{2}\\\\ -------------------------------\\\\ vertex~(2,-3)\qquad focus~\left(2+\frac{13}{2}~,~-3\right)[/tex]

check the picture below.

bear in mind that, because the leading term's coefficient is positive, namely for y², the parabola opens to the right, and "p" is positive, therefore the focus point will be to the right of the vertex.