Find the normal line to the curve \( y=\sqrt{x} \) at the point \( (a, \sqrt{a}) \). a. \( y=-2 \sqrt{a} x+\frac{1}{2} \sqrt{a} \). ob. None of these options. o c. \( y=\frac{1}{2 \sqrt{a}}+\frac{1}{2} \sqrt{a} \). od. \( y=x-2 a \sqrt{a}+\sqrt{a} \). e. \( y=-2 \sqrt{a} x-2 a \sqrt{a}+\sqrt{a} \).

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 \( y = \sqrt{x} \)**: \[ y = x^{1/2} \implies \frac{dy}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}}. \] 2. **Evaluate the derivative at \( x = a \)**: \[ \frac{dy}{dx} \bigg|_{x=a} = \frac{1}{2\sqrt{a}}. \] 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: \[ \text{slope of normal line} = -\frac{1}{\frac{1}{2\sqrt{a}}} = -2\sqrt{a}. \] 4. **Use the point-slope form to find the equation of the normal line**: The point-slope form of a line is given by: \[ y - y_1 = m(x - x_1), \] where \( (x_1, y_1) = (a, \sqrt{a}) \) and \( m = -2\sqrt{a} \). Plugging in these values: \[ y - \sqrt{a} = -2\sqrt{a}(x - a). \] 5. **Simplify the equation**: \[ y - \sqrt{a} = -2\sqrt{a}x + 2a\sqrt{a}. \] Rearranging gives: \[ y = -2\sqrt{a}x + 2a\sqrt{a} + \sqrt{a}. \] Combining the constant terms: \[ y = -2\sqrt{a}x + (2a + 1)\sqrt{a}. \] Now, let's compare this with the provided options: - a. \( y = -2\sqrt{a}x + \frac{1}{2}\sqrt{a} \) - b. None of these options. - c. \( y = \frac{1}{2\sqrt{a}} + \frac{1}{2}\sqrt{a} \) - d. \( y = x - 2a\sqrt{a} + \sqrt{a} \) - e. \( y = -2\sqrt{a}x - 2a\sqrt{a} + \sqrt{a} \) None of the options match the derived equation for the normal line. Therefore, the correct answer is: **b. None of these options.**