Answer:
A. Please find attached the drawing of the reflector showing the location of the focus created with Microsoft Excel
B. The width of the beam of light projected from the headlight's reflector is approximately 6.93 inches
Step-by-step explanation:
The given equation representing the cross section of the headlight reflector is y² = 6·x
The shape of the reflector = Parabolic
The general equation of a parabola is (y - k)² = 4·p·(x - h)
The focus of the parabola, Focus = (h + p, k)
By comparison to the general equation of a parabola, from the equation of the reflector, we have;
k = 0, h = 0, 4·p = 6
∴ p = 6/4 = 3/2 = 1.5
∴ The focus of the parabola = (1.5, 0)
A. Please find attached the graph of the equation of the cross-section of the reflector, showing the filament at the focus, created with Microsoft Excel
B. The depth of the reflector is x = 2 inches
A vertical or horizontal parabola has two x-values for a given y-value or two y-values for a given x-value respectively
For the given horizontal parabola, the width of a parabola at a point is the difference between the two y-values at the point
At x = 2, we have;
y² = 6 × x = 6 × 2 = 12
∴ y = ±√(12)
The two y-values at x = 2 are y₁ = √(12) and y₂ = -√(12)
The width of the parabola, w = y₁ - y₂
The width of the parabola at x = 2, w = √(12) - (-√(12))
∴ w = √(12) +√(12) = 4·√3
The width of the parabola at x = 2, w = 4·√3 inches
The width of the parabola at x = 2 = The width of the beam of light projected from the headlight's reflector = 4·√3 inches
By giving the answer as a decimal and rounded to the hundredths place, we have;
The width of the beam of light projected from the headlight's reflector ≈ 6.93 inches.
