(a)
\( x = -\frac{1}{k} \)

Curve: \( y = kx^2 – x \)
Slope: \( \frac{dy}{dx} = 2kx – 1 \)
At \( A_k\left(\frac{1}{k}, 0\right) \): \( m_{\text{tan}} = 2k \times \frac{1}{k} – 1 = 1 \)
Normal slope: \( m_{\text{normal}} = -1 \)
Normal equation: \( y = -x + \frac{1}{k} \)
Intersection: \( kx^2 – x = -x + \frac{1}{k} \)
Simplify: \( kx^2 – \frac{1}{k} = 0 \implies k^2 x^2 = 1 \implies x = \pm \frac{1}{k} \)
Since \( x = \frac{1}{k} \) is \( A_k \), \( B_k \) has \( x = -\frac{1}{k} \)

Result: The \( x \)-coordinate of \( B_k \) is \( -\frac{1}{k} \) [4]

(b)
Area = \( \frac{1}{6} \approx 0.333 \)

Curve for \( k = 2 \): \( y_1 = 2x^2 – x \)
Normal at \( A_2\left(\frac{1}{2}, 0\right) \): \( y = -x + \frac{1}{2} \)
Limits: \( x = 0 \) to \( x = \frac{1}{2} \) (normal above curve)
Area: \( \int_0^{\frac{1}{2}} \left( \left(\frac{1}{2} – x\right) – \left(2x^2 – x\right) \right) dx \)
Simplify: \( \int_0^{\frac{1}{2}} \left( \frac{1}{2} – 2x^2 \right) dx \)
Calculate: \( \left[ \frac{x}{2} – \frac{2x^3}{3} \right]_0^{\frac{1}{2}} = \frac{1}{4} – \frac{1}{12} = \frac{1}{6} \)

Result: Area = \( \frac{1}{6} \approx 0.333 \) [3]