A cross-section of a parabolic reflector and bulb is located at the focus is 16cm.find the equation of the parabola and the diameter of the opening (R,S),11cm from the vertex
Your reflector is just a parabola revolved about its axis.
Recall that the parabola
x^2 = 4py has
latus rectum = 2p
focus at (0,p)
vertex at (0,0)
So, your parabola has p=8, making its equation
x^2 = 32y
Now you want 2x where y=11
That will be 2√(32*11) = 8√22
Why did the math teacher bring a ladder to the parabolic reflector party?
Because he heard there would be some major equation solving going on, and he didn't want to miss out on the quadratic fun!
Now, let's get down to business. The equation of a parabola in vertex form is given by the equation:
y = a(x - h)^2 + k
In this case, the vertex of the parabola is at (h, k) = (0, 0), since it's located at the focus. We also know that the cross-section is 16cm away from the focus, which means the value of x at the vertex is 16cm. Plugging these values into the equation, we get:
y = ax^2
Now, to find the value of a, we can use the fact that the cross-section is located at (R, S) = (11cm, 16cm). Plugging these values into the equation, we get:
16cm = a(11cm)^2
Simplifying this equation, we find:
a = 16cm / (11cm)^2
Now that we have the value of a, we can substitute it back into the equation to get the final equation of the parabola. As for the diameter of the opening, we know that it is located 11cm from the vertex, so the distance between the two ends of the opening is twice that, which is 22cm.
So, the equation of the parabola is y = (16cm / (11cm)^2) x^2, and the diameter of the opening is 22cm. Hope that sheds some light on the situation!
To find the equation of the parabola, we need to determine the coordinates of the vertex and the focus.
Given that the cross-section of the parabolic reflector and bulb is located at the focus, we know that the focus is (0, 16 cm).
We are also given that the distance from the vertex to the point on the parabola (R, S) is 11 cm. This means that the vertex is 11 cm inside the parabola from the opening.
Let's assume that the vertex of the parabola is (0, 0). Since the distance from the vertex to the focus is 16 cm, we can determine the equation of the parabola using the formula:
y^2 = 4ax
Here, a is the distance from the vertex to the directrix.
Using the information given, we can find the value of a:
2a = 16
a = 8
Now, substituting the value of 'a' into the equation, we get:
y^2 = 32x
To find the diameter of the opening, we need to determine the width of the parabola at the opening. Since the vertex is 11 cm inside the parabola from the opening, the distance from the vertex to the opening is 11 cm.
Substituting x = 11 into the equation y^2 = 32x, we can solve for y:
y^2 = 32 * 11
y^2 = 352
y = √352
y ≈ 18.77 cm
Since the vertex is at the origin, the width at the opening is twice the y-value:
Width = 2y ≈ 2 * 18.77 cm ≈ 37.54 cm
Thus, the diameter of the opening is approximately 37.54 cm.
To find the equation of a parabola, in general, we need to know the focus (F) and the directrix (D). In this case, the focus is given as (R, S) = (16cm, 0cm), and the directrix is 11cm from the vertex.
The equation of a parabola in general form is given by:
(y - k)^2 = 4a(x - h)
where (h, k) is the vertex, and 'a' is the distance from the vertex to the focus or vertex to the directrix. In this case, the vertex is at (0cm, 0cm).
Since the directrix is 11cm from the vertex, we can say that 'a' is equal to half of the distance between the focus and the directrix. So, a = (11cm - 0cm)/2 = 5.5cm.
Now we can substitute the values of 'a' and the vertex into the general form equation to get the specific equation for this parabola:
(y - 0cm)^2 = 4 * 5.5cm * (x - 0cm)
Simplifying the equation gives:
y^2 = 22cm * x
Hence, the equation of the parabola is y^2 = 22x.
To find the diameter of the opening, we need to find the width of the parabola at the vertex. We know that the width at the vertex is equal to the distance between the focus and the directrix, which is 11cm in this case.
Since the focus is located at (16cm, 0cm) and the directrix is 11cm from the vertex, the width at the vertex is twice the x-coordinate of the focus, which is 2 * 16cm = 32cm.
Finally, the diameter of the opening is equal to the width at the vertex, so the diameter is 32cm.