1. Given Curve: \[ x^2 = 4y \] This is a parabola that opens upwards. 2. Find the Slope of the Tangent: Differentiate the equation of the curve with respect to \( x \): \[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4y) \Rightarrow 2x = 4 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2} \] The slope of the tangent at any point \( (x, y) \) on the curve is \( \frac{x}{2} \). 3. Slope of the Normal: The slope of the normal is the negative reciprocal of the slope of the tangent: \[ m_{\text{normal}} = -\frac{2}{x} \] 4. Equation of the Normal: The equation of the normal at a point \( (x, y) \) on the curve is: \[ y - y_1 = m_{\text{normal}}(x - x_1) \] Substituting \( y_1 = \frac{x^2}{4} \) and \( m_{\text{normal}} = -\frac{2}{x} \): \[ y - \frac{x^2}{4} = -\frac{2}{x}(x - x_1) \] Since the normal passes through \( (1, 2) \), substitute \( x = 1 \) and \( y = 2 \): \[ 2 - \frac{x^2}{4} = -\frac{2}{x}(1 - x) \] Simplify and solve for \( x \): \[ 2 - \frac{x^2}{4} = -\frac{2}{x} + 2 \] \[ -\frac{x^2}{4} = -\frac{2}{x} \] \[ \frac{x^2}{4} = \frac{2}{x} \] \[ x^3 = 8 \Rightarrow x = 2 \] Substituting \( x = 2 \) back into the equation of the normal: \[ y - \frac{2^2}{4} = -\frac{2}{2}(x - 2) \Rightarrow y - 1 = -1(x - 2) \Rightarrow y = -x + 3 \] 5. Distance of the Origin from the Normal: The distance \( D \) from the origin \( (0, 0) \) to the line \( y = -x + 3 \) is given by: \[ D = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} \] For the line \( y = -x + 3 \), rewrite it in standard form: \[ x + y - 3 = 0 \] Here, \( A = 1 \), \( B = 1 \), and \( C = -3 \). Substituting the coordinates of the origin: \[ D = \frac{|1 \cdot 0 + 1 \cdot 0 - 3|}{\sqrt{1^2 + 1^2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \] Therefore, the distance of the origin from the normal is: \[ \boxed{\frac{3}{\sqrt{2}}} \]