Find the normal line to the curve \( y=\sqrt{x} \) at the point \( (a, \sqrt{a}) \)

To find the normal line to the curve \( y = \sqrt{x} \) at the point \( (a, \sqrt{a}) \), we first need to determine the slope of the tangent line at that point. 1. **Find the derivative of the curve**: The derivative of \( y = \sqrt{x} \) can be calculated using the power rule: \[ y = x^{1/2} \implies \frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}. \] 2. **Evaluate the derivative at \( x = a \)**: The slope of the tangent line at the point \( (a, \sqrt{a}) \) is: \[ \frac{dy}{dx} \bigg|_{x=a} = \frac{1}{2\sqrt{a}}. \] 3. **Find the slope of the normal line**: The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is: \[ m_{\text{normal}} = -\frac{1}{\frac{1}{2\sqrt{a}}} = -2\sqrt{a}. \] 4. **Use the point-slope form to write the equation of the normal line**: The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) \) is the point on the line and \( m \) is the slope. Substituting \( (x_1, y_1) = (a, \sqrt{a}) \) and \( m = -2\sqrt{a} \), we have: \[ y - \sqrt{a} = -2\sqrt{a}(x - a). \] 5. **Simplify the equation**: Distributing the slope on the right side: \[ y - \sqrt{a} = -2\sqrt{a}x + 2a. \] Adding \( \sqrt{a} \) to both sides gives: \[ y = -2\sqrt{a}x + 2a + \sqrt{a}. \] Thus, the equation of the normal line to the curve \( y = \sqrt{x} \) at the point \( (a, \sqrt{a}) \) is: \[ y = -2\sqrt{a}x + (2a + \sqrt{a}). \]