Question 1164281
I already posted a solution of the same problem that was also posted as Question 1164605.
I will add context to relate concepts algebra, geometry, precalculus, and physics that are taught separately, and with little relation to history or to the real world.
 
A paraboloid is a bowl-shaped surface, like the surface swept by a parabola as it rotates around its axis of symmetry.
A cross section "cutting" through a plane containing the axis is a parabola.
If we place the x- and y-axes with the axis of the paraboloid along the y-axis and the origin at the apex of the paraboloid,
the cross section would be a parabola that we can represent by the familiar equation {{{y=ax^2}}}, and a graph like the one below:
{{{graph(350,220,-3.5,3.5,-2,2.4,0.2x^2,-1.25,1.25-sqrt(0.004-x^2))}}}
 
THE DEFINITION OF A PARABOLA is the set of points in a plane that are at equal distance from a point called the {{{blue(focus)}}} ,
and a line called the {{{green(directrix)}}}, as shown in the figure above.
That definition is about 1700 years old, but wise ancient Greeks knew parabolas some 2300 years ago, based on their knowledge of geometry and ratios.
Other civilizations had similarly wise men, developing mathematical concepts independently.
The idea of the coordinate plane came later, and it took longer for the standardization of the ways of representing and communicating math accepted across cultures that we use now.
When represented on an x-y coordinate plane, the directrix of a parabola can be any line, but to make equations easier, I chose to put the vertex of the parabola at the origin and to make the directrix parallel to the x-axis.
For that parabola, those choices give us an easy equation :  {{{y=ax^2}} .
With {{{a>0}}} and a focal distance distance {{{f}}} ,
the focus will be at point {{{F(f,0)}}} , and the directrix will be the line {{{y=-f}}} , at the same distance from  vertex {{{V(0,0)}}} .
Considering the point of the parabola with {{{y=f}}} , {{{P(x[F],f)}}} , with {{{x[F]>0}}} , we see that its distance to directrix {{{y=-f}}} is {{{2f}}}.
That must be the same distance to focus {{{F(0,f)}}} , so {{{x[F]=2f}}} .
The point {{{P(2f,f)}}} , being a point of the parabola with equation {{{y=ax^2}}} ,
it must satisfy that equation, so
{{{f=a(2f)^2}}}-->{{{f=4af^2}}}-->{{{4af^2-f=0}}}-->{{{(4af-1)f=0}}}-->{{{4af=1}}}-->{{{highlight(a=1/4f)}}}<-->{{{highlight(f=1/4a)}}}
The relations highlighted are not formulas to memorize. They will be remembered as long as they are being used often, but can be easily discovered/rediscovered as needed.
 
BACK TO THE PROBLEM:
As we get further away from the apex/vertex, a paraboloid (and a parabola) go on widening forever, but the reflecting surface of the flashlight is just a finite piece of a paraboloid with a depth of 1.8 inches and a diameter of 6 inches at the open end.
The corresponding graph would follow the parabola, but end at the points (-3,1.8) and (3,1.8).
The endpoints with {{{x=" " +- 3}}} and {{{y=1.8}}} would satisfy {{{y=ax^2}}}, so
{{{1.8=a*( " " +- 3)^2}}} , {{{1.8=9a}}} and {{{a=1.8/9=0.2}}}
From there, if we like to remember formulas, we could use the one that says that
the focal distance of a parabola with the equation {{{y=ax^2}}}
is {{{f=1/4a}}} to calculate {{{f=1/(4*0.2)=1/0.8=highlight(1.25)}}}
that is the the distance from the focus of the parabola to its vertex,
from the focus of the paraboloid to its apex,
and the distance we need between the light source to the center of the bottom (apex) of the paraboloid-shaped reflective "bowl".  
 
PHYSICS OF REFLECTION ON PARABOLIC SURFACES:
Reflection is simple. Smooth surfaces can be planar, or have a tangent plane, and there is a normal direction perpendicular to that plane.
Any ray aimed at a reflecting surface at an angle with respect to the normal direction, is reflected with the same angle to the opposite side.
The parabola and parabolic reflecting surfaces were known to wise ancient Greeks about 2300 years ago.
The parabola, reflection, and the properties of paraboloid reflecting surfaces were known to wise ancient Greeks about 2300 years ago.
They knew that they could use a paraboloid reflecting surface to reorganize light coming in all directions from a point source located at the focus of the paraboloid into a beam of rays all going in the direction of the paraboloid axis of symmetry.
They also knew that they could use a paraboloid reflecting surface to concentrate a beam of rays parallel to the paraboloid axis of symmetry into a point at the focus of the paraboloid.

How does that work?
{{{drawing(350,260,-3.5,3.5,-2,3.2,
graph(350,260,-3.5,3.5,-2,3.2,-1.25x+3.3,-1.25,0.8(x-1),0.2x^2),
triangle(1,1.025,0,1.25,2,0.8),red(arrow(0,1.25,2,0.8)),
circle(0,1.25,0.05),locate(2.2,0.95,B),locate(2.05,3,H),
locate(0.05,1.55,blue(F)),triangle(2,-3.5,2,0,2,3.2),
red(arrow(2,0.8,2,2.85)),arc(2,0.8,1.1,1.1,192,231),arc(2,0.8,1,1,231,270),
locate(0.3,-0.5,blue(tangent)),locate(0.5,3,red(normal)),
locate(-3,-1.25,green(directrix))
)}}} Ray {{{red(FB)}}} reflects as ray {{{red(BH)}}} ,or
ray {{{red(HB)}}} reflects as ray {{{red(BF)}}} {{{drawing(350,260,-3.5,3.5,-2,3.2,
graph(350,260,-3.5,3.5,-2,3.2,-1.25x+3.3,-1.25,0.8(x-1),0.2x^2),
triangle(1,1.025,0,1.25,2,0.8),red(arrow(2,0.8,0,1.25)),
circle(0,1.25,0.05),locate(2.2,0.95,B),locate(2.05,3,H),
locate(0.05,1.55,blue(F)),triangle(2,-3.5,2,0,2,3.2),
red(arrow(2,2.85,2,0.8)),arc(2,0.8,1.1,1.1,192,231),arc(2,0.8,1,1,231,270),
locate(0.3,-0.5,blue(tangent)),locate(0.5,3,red(normal)),
locate(-3,-1.25,green(directrix))
)}}} .