Question

The area bounded by the parabola $y^2 = 4ax$ and its latus-rectum $x = a$ is

• A. $\frac{8}{3} a^2$ sq. units
• B. $\frac{2}{3} a^2$ sq. units
• C. $\frac{4}{3} a^2$ sq. units
• D. $8 a^2$ sq. units

Hint:

This is a classic standard area tracking formula in integral calculus! The area bounded by a standard parabola $y^2 = 4ax$ and its vertical latus-rectum line $x=a$ is always globally fixed as exactly $\frac{8}{3}a^2$. Memorizing this specific standard result can save you over a minute of manual integration during exams.

Answer 1

The Correct Option is A

Step 1: Picture the region.
The parabola $y^2 = 4ax$ opens rightward, and the latus-rectum is the vertical line $x = a$ through the focus. The bounded region lies between the curve and this line.
Step 2: Use symmetry.
The region is symmetric about the X-axis, so total area $= 2 \times$ (area of the upper half).
Step 3: Choose horizontal strips, integrate over $y$.
For the upper half, the curve gives $x = \dfrac{y^2}{4a}$ and the line is $x = a$. As $x$ ranges $0$ to $a$, $y$ ranges $0$ to $2a$. The width of a strip is $a - \dfrac{y^2}{4a}$.
Step 4: Set up the integral.
Upper area $= \displaystyle\int_0^{2a}\left(a - \dfrac{y^2}{4a}\right)dy$.
Step 5: Integrate.
$= \left[ay - \dfrac{y^3}{12a}\right]_0^{2a} = a(2a) - \dfrac{(2a)^3}{12a} = 2a^2 - \dfrac{8a^3}{12a} = 2a^2 - \dfrac{2a^2}{3} = \dfrac{4a^2}{3}$.
Step 6: Double for full area.
Total area $= 2 \times \dfrac{4a^2}{3} = \dfrac{8}{3}a^2$ sq. units.
\[ \boxed{\dfrac{8}{3}a^2 \text{ sq. units}} \]