I asked this question yesterday, but I wasn't clear about the response. Could someone please explain it again?
A searchlight is shaped like a paraboloid of revolution. If the light source is located 2 feet from the base along the axis of symmetry and the opening is 5 feet across, how deep should the search light be?
I don't see how you can find an equation with only that information.
The focal length is 2 feet. That is the didtanbce from the light-producing arc to the bottom of the mirror. That focal distance equals 1/4a, where a is the constant in the equation y = ax^2 of the equation of the parabola that generates the paraboloidal surface of the searchlight mirror.
The problem should have made clear why a paraboloidal mirror is used and why the light source is placed at the focus. Paraboloids produce the best parallel "collimated" beam of light. Telescope mirrors are paraboloids for the same reason.
Therefore
1/(4a) = 2
That equation comes from basic analytic geometry of parabolas, which you should perhaps review.
4a = 1/2
a = 1/8
y = x^2/8 is the equation of the parabola, if y and x are in feet.
At the top of the parabola, x = 2.5 ft since it is 5 feet across.
Therefore the depth of the parabola is
(2.5)^2/8 = 0.781 feet.
To find the depth of the searchlight, we can start by visualizing a cross-section of the paraboloid shape. Since the searchlight is shaped like a paraboloid of revolution, we can imagine slicing through it vertically to create a cross-section.
In this cross-section, the opening of the searchlight forms a semi-circle with a diameter of 5 feet. We know that the light source is located 2 feet from the base along the axis of symmetry.
To find the depth of the searchlight, we need to determine the height of the paraboloid. This height refers to the distance between the vertex (the deepest point of the parabola) and the base of the searchlight.
Now, let's consider the coordinates of the cross-section. Let's represent the semi-circle's center (the point on the axis of symmetry) as (0, 0).
Since the diameter of the semi-circle is 5 feet, the leftmost point on the semi-circle is at (-2.5, 0), and the rightmost point is at (2.5, 0). Additionally, the highest point of the semi-circle is at (0, r), where r represents the radius of the semi-circle.
Now, let's consider the equation of a parabola in its vertex form: y = a(x - h)^2 + k, where (h, k) represents the vertex.
Since the semi-circle's vertex is at (0, r), we can write the equation as: y = a(x - 0)^2 + r.
Now, we substitute the coordinates of the leftmost point (-2.5, 0) into the equation: 0 = a(-2.5 - 0)^2 + r.
Next, let's substitute the coordinates of the rightmost point (2.5, 0) into the equation: 0 = a(2.5 - 0)^2 + r.
We can solve these two equations simultaneously to find the values of a and r.
From the first equation, we have: 0 = 6.25a + r.
From the second equation, we have: 0 = 6.25a + r.
Since the leftmost and rightmost points have the same y-coordinate (0), the values of a and r must be equal.
Therefore, we can rewrite the equation: 0 = 6.25a + a.
Simplifying, we have: 0 = 7.25a.
To find a, we divide both sides of the equation by 7.25: a = 0.
This means that the value of a is 0, and since a represents the coefficient of the quadratic term (x^2) in the parabola equation, we can conclude that the paraboloid is actually a flat surface in this case.
Therefore, there is no depth to the searchlight. It is essentially a flat reflector in the shape of a semi-circle.