Area under parabolas
For y² = 4ax, the upper half is y = 2√(ax). The region cut off by the latus rectum x = a has area 8a²/3.
The four standard parabolas (a > 0)
| Equation | Opens towards | Symmetric about | Half to integrate |
|---|---|---|---|
| y² = 4ax | the right | x-axis | y = 2√(ax), x ≥ 0 |
| y² = −4ax | the left | x-axis | y = 2√(−ax), x ≤ 0 |
| x² = 4ay | upwards | y-axis | x = 2√(ay), or y = x²/(4a) |
| x² = −4ay | downwards | y-axis | x = 2√(−ay), or y = −x²/(4a) |
Area cut off by the latus rectum of y² = 4ax
- 1The latus rectum is the line x = a. The region is symmetric about the x-axis.
- 2A = 2 ∫ y dx from 0 to a = 2 ∫ 2√(ax) dx from 0 to a
- 3= 4√a × ∫ √x dx from 0 to a = 4√a × (2/3)[x√x] from 0 to a
- 4★= 4√a × (2/3) × a√a = (8/3)a²
★ key step
Example 1: vertical strips
Find the area of the region in the first quadrant bounded by the parabola y² = 4x, the x-axis and the lines x = 1 and x = 4.
- In the first quadrant, y = 2√x.
- A = ∫ 2√x dx from 1 to 4
- = 2 × (2/3)[x√x] from 1 to 4
- = (4/3)(8 − 1)
- = 28/3
Area = 28/3 sq units.
Example 2: horizontal strips
Find the area of the region in the first quadrant bounded by the parabola x² = 4y, the y-axis and the lines y = 1 and y = 9.
- The boundaries are the y-axis and two horizontal lines, so use horizontal strips.
- In the first quadrant, x = 2√y.
- A = ∫ x dy from 1 to 9 = ∫ 2√y dy from 1 to 9
- = 2 × (2/3)[y√y] from 1 to 9
- = (4/3)(27 − 1) = (4/3) × 26 = 104/3
Area = 104/3 sq units.
Example 3: both halves of the parabola
Find the area of the region bounded by the parabola y² = 9x and the line x = 4.
- The region runs from x = 0 to x = 4 and is symmetric about the x-axis.
- Upper half: y = √(9x) = 3√x.
- A = 2 ∫ 3√x dx from 0 to 4 = 6 × (2/3)[x√x] from 0 to 4
- = 4 × 8 = 32
Area = 32 sq units.
How to draw the parabola
- Draw the axes and mark the vertex at O.
- For y² = 4ax, plot a few points, e.g. for y² = 9x: (1, 3), (1, −3), (4, 6) and (4, −6). Draw a smooth curve opening to the right.
- For x² = 4ay the curve opens upwards, e.g. for x² = 4y: (±2, 1) and (±6, 9).
- Draw the boundary lines, mark where they meet the curve, and shade the region.
- Keep the curve smooth and rounded at the vertex, never pointed.
Satellite dishes and scooter headlights have parabolic shapes. The area under a parabola also gives the face area of a parabolic arch in a bridge or a temple gateway.
1 marks: “Write the area of the region bounded by the parabola y² = 4ax and its latus rectum.” (8/3)a² sq units.
3 marks: “Find the area of the region bounded by the parabola y² = 16x and its latus rectum.” 4a = 16, so a = 4 and the latus rectum is x = 4. A = 2 ∫ 4√x dx from 0 to 4 = 8 × (2/3) × 8 = 128/3 sq units.
3 marks: “Find the area of the region in the first quadrant bounded by the parabola x² = 6y, the x-axis and the line x = 3.” y = x²/6. A = ∫ x²/6 dx from 0 to 3 = [x³/18] from 0 to 3 = 27/18 = 3/2 sq units.
Latus rectum area = 8a²/3. Check: for y² = 4x (a = 1) it is 8/3.