1. Write the parabola in standard form

Given $y^2=x-2 \implies x=y^2+2$ This is a right-opening parabola with vertex at $(2,0)$.

The point of contact $B$ lies in the first quadrant, so let $B=(2+t^2,t), \quad t>0.$

2. Equation of tangent to the parabola at parameter $t$

For the parabola $y^2=4a(x-h),$ with $4a=1$ and $h=2$, the tangent at $(2+t^2,t)$ can be written using differentiation.

From $y^2=x-2,$ differentiate: $2y\frac{dy}{dx}=1 \implies \frac{dy}{dx}=\frac{1}{2y}.$ So at $B$, slope of tangent is $m=\frac{1}{2t}.$

Hence tangent at $(2+t^2,t)$ is $y-t=\frac{1}{2t}\big(x-(2+t^2)\big).$

3. Use the fact that tangent passes through $A(-2,0)$

Substitute $(-2,0)$ into the tangent equation: $0-t=\frac{1}{2t}\big(-2-(2+t^2)\big).$ So, $-t=\frac{-4-t^2}{2t}.$ Multiply by $2t$: $-2t^2=-4-t^2$ $-t^2=-4$ $t^2=4.$ Since $t>0$, $t=2.$

Therefore, $B=(2+4,2)=(6,2).$

4. Find equation of line $AB$

Slope of $AB$: $m=\frac{2-0}{6-(-2)}=\frac{2}{8}=\frac14.$ Hence equation is $y=\frac14(x+2)=\frac{x}{4}+\frac12.$ Or in terms of $x$ as a function of $y$: $x=4y-2.$

5. Identify the bounded region

The region is bounded by:

• the line segment $AB$
• the parabola from $(2,0)$ to $(6,2)$
• the $x$-axis from $(-2,0)$ to $(2,0)$

It is convenient to integrate with respect to $y$ from $y=0$ to $y=2$.

For a fixed $y\in[0,2]$:

• left boundary is the line: $x=4y-2$
• right boundary is the parabola: $x=y^2+2$

So area is $\text{Area}=\int_0^2 \big[(y^2+2)-(4y-2)\big] \,dy.$

Simplify: $\text{Area}=\int_0^2 (y^2-4y+4)\,dy=\int_0^2 (y-2)^2\,dy.$

6. Evaluate the integral

$\text{Area}=\int_0^2 (y^2-4y+4)\,dy$ $=\left[\frac{y^3}{3}-2y^2+4y\right]_0^2$ $=\left(\frac{8}{3}-8+8\right)-0$ $=\frac{8}{3}.$

7. Check options

The area is $\boxed{\frac{8}{3}}$ So the correct option is C.

8. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.