Application of Derivatives — Class 12 Maths Solution

We have,
${y^2} = 4x$ …(i)

Differentiating (i) w.r.t. $x$, we get $2y\cfrac{{dy}}{{dx}} = 4 \Rightarrow \cfrac{{dy}}{{dx}} = \cfrac{4}{{2y}} = \cfrac{2}{y}$

Therefore, Slope of tangent to (i) at $(1,\;2)$ is ${\left( {\cfrac{{dy}}{{dx}}} \right)_{(1,2)}} = 1$

$\Rightarrow$ Slope of normal to (i) at $(1,\;2) = - 1$

Therefore equation of normal to (i) at $(1,\;2)$ is
$(y - 2) = - 1(x - 1)$

$\Rightarrow x + y - 3 = 0.$