NIMCET 2014 Question Paper with solutions

We have,
${\mathit{x}}^2$ = 4y
On differentiating 2x = 4 $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$
or $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = x/2
Slope of normal = -2 / x
Let the common point of curve and normal be (to, k)
Then,
${\mathit{h}}^2$ = 4k ... (i)
Equation of normal $\frac{\mathit{y}-\mathit{k}}{\mathit{x}-\mathit{h}}$ = $\frac{-2}{\mathit{h}}$
Normal passes through (1,2),
∴ $\frac{2-\mathit{k}}{1-\mathit{h}}$ = $\frac{-2}{\mathit{h}}$
2h- kh = -2+2h
From Eq. (i), 2h - $\frac{{\mathit{h}}^3}{4}$ = - 2 + 2h
or $\frac{{\mathit{h}}^3}{4}$ = 2 ⇒ h = 2
k = 1
Equation of normal x + y - 3 = 0
Now, distance form origin (0, 0)
= $\left|\frac{0.0-4}{\sqrt{1+1}}\right|$ = $\left|\frac{-2}{\sqrt{\left(2\right)}}\right|$ = $\sqrt{2}$ units