Question

Find the area of the region enclosed by the parabola \( y^2 = 4ax \) and its latus rectum.

Hint:

Utilizing symmetry is a common strategy to simplify area calculations. Calculating the area in one quadrant and multiplying is often easier than dealing with positive and negative roots of y.

Answer 1

Step 1: Understanding the Concept:
The parabola \( y^2 = 4ax \) opens to the right with its vertex at the origin (0, 0). Its latus rectum is a line segment perpendicular to the axis of symmetry (the x-axis), passing through the focus at \( (a, 0) \). The equation of the line containing the latus rectum is therefore \( x = a \). The area is bounded by the parabola and this line.
Step 2: Key Formula or Approach:
The area of a region bounded by a curve \( y = f(x) \), the x-axis, and the lines \( x = c \) and \( x = d \) is given by \( \int_c^d f(x) dx \).
Since the parabola is symmetric about the x-axis, we can find the area in the first quadrant (from \( x=0 \) to \( x=a \)) and multiply it by 2.
Step 3: Detailed Explanation or Calculation:
From \( y^2 = 4ax \), we get \( y = \pm\sqrt{4ax} = \pm 2\sqrt{a}\sqrt{x} \).
For the area in the first quadrant, we take \( y = 2\sqrt{a}\sqrt{x} \).
The limits of integration are from the vertex \( x=0 \) to the latus rectum \( x=a \).
The area in the first quadrant is:
\[ A_1 = \int_0^a 2\sqrt{a}\sqrt{x} \, dx = 2\sqrt{a} \int_0^a x^{1/2} \, dx \] \[ A_1 = 2\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_0^a = 2\sqrt{a} \left[ \frac{2}{3}x^{3/2} \right]_0^a \] \[ A_1 = 2\sqrt{a} \left( \frac{2}{3}a^{3/2} - 0 \right) = \frac{4\sqrt{a}}{3}a^{3/2} = \frac{4}{3}a^{1/2}a^{3/2} = \frac{4}{3}a^2 \] The total area is twice the area in the first quadrant:
\[ A = 2 \times A_1 = 2 \times \frac{4}{3}a^2 = \frac{8}{3}a^2 \] Step 4: Final Answer:
The area of the region is \( \frac{8}{3}a^2 \) square units.