Area bounded between the parabola y² = 4ax and its latus rectum is

The correct option is 8/3 a² sq. units.

Let us understand how to find the area bounded by the parabola $y^2=4ax$ and its latus rectum step-by-step.

Step 1: Identify the curve and the boundary lines.
The equation of the parabola is given by:

$y^2=4ax$

This is a standard rightward-opening parabola with its vertex at the origin $(0,0)$.
The focus of this parabola is at the point $(a,0)$.
The latus rectum is the chord passing through the focus and perpendicular to the axis of the parabola (which is the x-axis, $y=0$). Therefore, the equation of the latus rectum is the vertical line:

$x=a$

Step 2: Understand the symmetry.
Since the power of $y$ in the equation of the parabola is even ($y^2$), the curve is symmetrical about the x-axis.
Thus, the total area bounded between the parabola and the latus rectum is twice the area of the region lying above the x-axis (in the first quadrant) from $x=0$ to $x=a$.
For the portion of the curve in the first quadrant, we take the positive square root:

$y=2\sqrt{a}\sqrt{x}$

Step 3: Set up the definite integral for the area.
The total area $A$ is given by:

$A=2{\int }_0^{a}ydx$

Substituting $y=2\sqrt{a}{x}^{1/2}$ into the integral:

$A=2{\int }_0^{a}2\sqrt{a}{x}^{1/2}dx$

$A=4\sqrt{a}{\int }_0^a{x}^{1/2}dx$

Step 4: Evaluate the integral.
Using the power rule for integration, $\int x^{n}dx=\frac{x^{n+1}}{n+1}$:

${\int }_0^a{x}^{1/2}dx={[\frac{x^{3/2}}{3/2}}_0^a=\frac{2}{3}{\left[x^{3/2}\right]}_0^a$

Applying the upper and lower limits:

$\frac{2}{3}(a^{3/2}-0)=\frac{2}{3}{a}^{3/2}$

Step 5: Calculate the final area.
Substitute this back into our expression for $A$:

$A=4\sqrt{a}\left(\frac{2}{3}{a}^{3/2}\right)$

$A=\frac{8}{3}{a}^{1/2}{a}^{3/2}$

$A=\frac{8}{3}{a}^2\text{ sq. units}$

Therefore, the area bounded between the parabola and its latus rectum is indeed 8/3 a² sq. units.