Tangents and Normals — MATHS STD 12 Science — Question

$x^2 = 4y$
$2\text{x}=4\frac{\text{dy}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2}$
$\Big(\frac{\text{dy}}{\text{dx}}\Big)_{(1,2)}=\frac{1}{2}=\text{m}$
Slope of normal $=\frac{-1}{\frac{\text{dy}}{\text{dx}}}=\frac{-1}{\frac{\text{x}}{\text{2}}}=\frac{-2}{\text{x}}$
Let $(X, Y)$ be the point where normal and curve intersect
$\therefore $ Slope of normal at $(X, Y) =\frac{-2}{\text{X}}$
Equation of normal passing through $(X, Y) $ with slope $\frac{-2}{\text{X}}$ is
$\text{y}-\text{y}_1=\text{m}(\text{x}-\text{x}_1)$
$\text{y}-\text{Y}=\frac{-2}{\text{h}}(\text{x}-\text{X})$
Since normal passes through $(1, 2)$ it will satisfy its equation
$\text{Y}=2+\frac{-2}{\text{h}}(\text{1}-\text{X})\ \dots(1)$
Since $(X, Y)$ lies on the curve $x^2 = 4y$
$X^2 = 4Y .......(2)$
Using $(1)$ and $(2)$
$\Rightarrow\ 2+\frac{-2}{\text{h}}(\text{1}-\text{X})=\frac{\text{X}^2}{4}$
$2+\frac{2}{\text{X}}-2=\frac{\text{X}^2}{4}$
$\frac{2}{\text{X}}=\frac{\text{X}^2}{4}$
$\text{X}^3=8$
$\text{X}=2$
Putting $X = 2$ in $(2)$
$\text{Y}=\frac{\text{X}^2}{4}=\frac{(2)^2}{4}=\frac{4}{4}=1$
Hence, $X = 2, Y = 1$
$\text{y}-\text{Y}=\frac{-2(\text{x}-\text{X})}{\text{X}}$
$\text{y}-\text{1}=\frac{-2(\text{x}-\text{2})}{\text{2}}$
$\text{y}-1=-1(\text{x}-2)$
$\text{y}-1=-\text{x}+2$
$\text{x}+\text{y}=2+1$
$\text{x}+\text{y}=3$