find the area of the region bounded by the parabola y^2= 16x and its latus rectum
To find the area of the region bounded by the parabola y^2 = 16x and its latus rectum, we need to determine the coordinates of the points where the parabola intersects the latus rectum.
The latus rectum of a parabola is a line segment perpendicular to the axis of symmetry and passing through the focus of the parabola. For a parabola given by the equation y^2 = 4ax, the latus rectum is 4a units long.
In this case, we have y^2 = 16x, so the coefficient 'a' is 4. Therefore, the length of the latus rectum is 4a = 4 * 4 = 16 units.
The focus of the parabola is located at (a, 0). Substituting a = 4 into the equation, we find the focus at (4, 0).
To find the points of intersection between the parabola and its latus rectum, let's set y^2 = 16x and y = 16.
(16)^2 = 16x
256 = 16x
x = 16
So the point of intersection between the parabola and its latus rectum is (16, 16).
Now, we can find the area of the region bounded by the parabola and its latus rectum. This area is the area under the parabola between x = 0 and x = 16.
To find the area, we integrate the function y^2 = 16x with respect to x from 0 to 16:
Area = ∫[0, 16] y^2 dx
= ∫[0, 16] 16x dx
= 16 ∫[0, 16] x dx
= 16 * (x^2/2) |[0, 16]
= 16 * (16^2/2) - 16 * (0^2/2)
= 16 * (256/2)
= 16 * 128
= 2048 square units
Therefore, the area of the region bounded by the parabola y^2 = 16x and its latus rectum is 2048 square units.
To find the area of the region bounded by the parabola y^2 = 16x and its latus rectum, we first need to understand what a latus rectum is.
The latus rectum of a parabola is a line segment perpendicular to the axis of symmetry and passing through the focus of the parabola. For a parabola with the equation y^2 = 4ax, where 'a' is a constant, the length of the latus rectum is given by 4a.
In our case, the equation of the parabola is y^2 = 16x. Comparing it with the standard form, we see that 4a = 16, which means a = 4.
Now, let's find the coordinates of the focus of the parabola. The focus lies at the point (a, 0) in standard form. So, for our parabola, the focus is located at (4, 0).
Next, we need to determine the equation of the latus rectum. Since the latus rectum passes through the focus (4, 0) and is perpendicular to the x-axis, its equation is given by x = a/2. Substituting the value of 'a' as 4, we get x = 4/2 = 2.
Now we have the boundaries of the region bounded by the parabola and its latus rectum, which are x = 0 and x = 2.
To find the area of the region, we integrate the equation of the parabola with respect to x over the given boundaries. Using the equation y^2 = 16x, we can rewrite it as y = ± √(16x).
Now we integrate the equation √(16x) with respect to x from 0 to 2. Note that we only consider the positive square root since the region is bounded above by the parabola.
∫[0 to 2] √(16x) dx = [2/3 * (16x)^(3/2)] from 0 to 2 = [2/3 * (16*2)^(3/2)] - [2/3 * (16*0)^(3/2)]
Simplifying further:
= [2/3 * 32^(3/2)] - [2/3 * 0^(3/2)]
= [2/3 * 32^(3/2)]
= [2/3 * 32 * √32]
= 2/3 * 32 * 4
= 2 * 32/3 * 4
= 256/3
So, the area of the region bounded by the parabola y^2 = 16x and its latus rectum is 256/3 square units.