Question

Find the area of the region bounded by the parabola \( y^2 = 16x \) and its latus rectum.

Hint:

When calculating areas for parabolas, use the formula for the area bounded by the curve and the axis, and integrate the equation of the parabola.

Answer 1

Step 1: Recall the equation of the parabola.
The given equation is \( y^2 = 16x \), which is the standard form of a parabola opening to the right with vertex at the origin.

Step 2: Find the focus and latus rectum.
For the parabola \( y^2 = 4ax \), we have \( 4a = 16 \), so \( a = 4 \). The focus is at \( (4, 0) \), and the length of the latus rectum is \( 4a = 16 \).

Step 3: Set up the integral for the area.
The area of the region bounded by the parabola and its latus rectum can be found by integrating the curve from \( x = 0 \) to \( x = 4 \) (the x-coordinate of the focus): \[ A = 2 \int_0^4 \sqrt{16x} \, dx = 2 \int_0^4 4\sqrt{x} \, dx \]

Step 4: Solve the integral.
Now, solve the integral: \[ A = 8 \int_0^4 \sqrt{x} \, dx = 8 \left[ \frac{2}{3} x^{3/2} \right]_0^4 = 8 \times \frac{2}{3} \left( 4^{3/2} - 0 \right) \] \[ = 8 \times \frac{2}{3} \times 8 = \frac{128}{3} \]

Step 5: Conclude the solution.
Thus, the area of the region is \( \frac{128}{3} \) square units.

Final Answer: \[ \boxed{\frac{128}{3}} \]