Application of Derivatives — Class 12 Maths Solution

Option A is correct

We have ${x^2} = 4y$ …(i)

Let $({x_1},\;{y_1})$ be any point on (i),

Therefore, $x_1^2 = 4{y_1}$ …(ii)

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

So, slope of the normal at $({x_1},\;{y_1}) = - \cfrac{2}{{{x_1}}}$

Therefore the equation of normal is $y - {y_1} = - \cfrac{2}{{{x_1}}}(x - {x_1})$ …(iii)

Since it passes through $(1,\;2)$, Therefore, $2 - {y_1} = - \cfrac{2}{{{x_1}}}(1 - {x_1})$

$\Rightarrow 2{x_1} - {x_1}{y_1} = - 2 + 2{x_1} \Rightarrow {x_1}{y_1} = 2 \Rightarrow {y_1} = \cfrac{2}{{{x_1}}}$

Putting ${y_1} = \cfrac{2}{{{x_1}}}$ in

(ii),

${x_1}^2 = 4 \cdot \cfrac{2}{{{x_1}}} \Rightarrow x_1^3 = 8 \Rightarrow {x_1} = 2$

Therefore, ${y_1} = \cfrac{2}{2} = 1.$

Putting ${x_1} = 2,{y_1} = 1$ in (iii), then the equation of the normal is $y - 1 = - \cfrac{2}{2}(x - 2) \Rightarrow y - 1 = - x + 2 = x + y = 3.$