If (h, k) be the point to which the origin has to be shifted in order to get the transformed equation of $y^2 – 4x + 6y + 17 = 0$ as $y^2 = 4ax$, then $h^2 + k^2 =$

Given Original Equation ${\mathrm{y}}^2-4\mathrm{x}+6\mathrm{y}+17=0$

Sub $\mathrm{x}=X+\mathrm{h}, \mathrm{y}=Y+\mathrm{k}$

Transformed equation is 

$$\begin{gathered} {(Y+\mathrm{k})}^2-4(X+\mathrm{h})+6(Y+\mathrm{k})+17=0 \\ \Rightarrow Y^2+{\mathrm{k}}^2+2kY-4X-4\mathrm{h}+6Y+6\mathrm{k}+17=0 \\ \Rightarrow Y^2+(2\mathrm{k}+6)Y-4X-4\mathrm{h}+6\mathrm{k}+{\mathrm{k}}^2+17=0 \\ \mathrm{Compare} \mathrm{with} {\mathrm{y}}^2=4\mathrm{ax} \\ \mathrm{we} \mathrm{get} 2\mathrm{k}+6=0 \mathrm{and} -4\mathrm{h}+6\mathrm{k}+{\mathrm{k}}^2+17=0 -\left(1\right) \\ \Rightarrow \mathrm{k}=-3 \\ \mathrm{sub} \mathrm{k}=-3 \mathrm{in} \left(1\right) \\ -4\mathrm{h}+6(-3)+{(-3)}^2+17=0 \\ \Rightarrow -4\mathrm{h}-18+9+17=0 \\ \Rightarrow -4\mathrm{h}=-8 \\ \Rightarrow \mathrm{h}=2 \\ \mathrm{Now} {\mathrm{h}}^2+{\mathrm{k}}^2=4+9=13 \end{gathered}$$