NIMCET 2015 Question Paper with solutions

Let $\left({\mathit{h}}_1,\mathit{k}\right)$ be the mid-point of chord of parabola ${\mathit{y}}^2=4\mathit{x}$.
∴ Equation of chord is $\mathit{y}\mathit{k}-2\left(\mathit{x}+\mathit{h}\right)={\mathit{k}}^2-4\mathit{h}$
[∵ Equation of chord of parabola when mid-point (x, y) is $\mathit{T}={\mathit{S}}_1$]
$\Rightarrow \mathit{y}\mathit{k}-2\mathit{x}-2\mathit{h}={\mathit{k}}^2-4\mathit{h}$...(i)
Chord are drawn through vertex
$\therefore \mathit{h}=\frac{\mathit{x}}{2}$ and $\mathit{k}=\frac{\mathit{y}}{2}$
$\Rightarrow 2\mathit{h}=\mathit{x}$ and $2\mathit{k}=\mathit{y}$
Putting the values of x and y in Eq. (i), we get
$2{\mathit{k}}^2-4\mathit{h}-2\mathit{h}={\mathit{k}}^2-4\mathit{h}{\mathit{k}}^2=2\mathit{h}$
∴ Locus of the mid-point of the parabola is ${\mathit{y}}^2=2\mathit{x}$