Find the equation of the normal at to the curve .

Given curve: $x^2 = 4y \implies y = \frac{x^2}{4}$. $\frac{dy}{dx} = \frac{x}{2}$. At $x = 1$, slope of tangent is $m_T = \frac{1}{2}$, so slope of normal is $m_N = -2$. Using the point $(1, 2)$: $y - 2 = -2(x - 1) \implies 2x + y = 4$ (or $2x + y - 4 = 0$). Note: The given point $(1, 2)$ does not lie on the curve $x^2 = 4y$. If the question intended the normal passing through $(1, 2)$, the normal with foot $(2, 1)$ is $x + y = 3$ (and through $(0, 0)$ is $x = 0$).