The normal to the curve

The equation of the given curve is $x^{2}=4 y$.

Differentiating with respect to x, we have:

$2 x=4 \cdot \frac{d y}{d x}$

$\Rightarrow \frac{d y}{d x}=\frac{x}{2}$

The slope of the normal to the given curve at point (h, k) is given by,

$\frac{-1}{\left.\frac{d y}{d x}\right]_{(h, k)}}=-\frac{2}{h}$

∴Equation of the normal at point (h, k) is given as:

$y-k=\frac{-2}{h}(x-h)$

Now, it is given that the normal passes through the point (1, 2).

Therefore, we have:

$2-k=\frac{-2}{h}(1-h)$ or $k=2+\frac{2}{h}(1-h)$

Since $(h, k)$ lies on the curve $x^{2}=4 y$, we have $h^{2}=4 k$.

$\Rightarrow k=\frac{h^{2}}{4}$

From equation (i), we have:

$\frac{h^{2}}{4}=2+\frac{2}{h}(1-h)$

$\Rightarrow \frac{h^{3}}{4}=2 h+2-2 h=2$

$\Rightarrow h^{3}=8$

$\Rightarrow h=2$

$\therefore k=\frac{h^{2}}{4} \Rightarrow k=1$

Hence, the equation of the normal is given as:

$\Rightarrow y-1=\frac{-2}{2}(x-2)$

$\Rightarrow y-1=-(x-2)$

$\Rightarrow x+y=3$

The correct answer is A.