Particle going on a field trip

Let us first find the equation of the electric field line which passes through $(1,1)$ .As $V={x}^{2}+y$ Then electric field $E$ at any point $(x,y)$ is given by $\overrightarrow { E } =-\frac { \partial }{ \partial x } (V)\hat { i } -\frac { \partial }{ \partial y } (V)\hat { j } -\frac { \partial }{ \partial z } (V)\hat { k }$ That is $\overrightarrow { E } =-2x\hat { i } -1\hat { j }$ Now let the equation of electric field line be $y=f\left( x \right)$ . So the tangent to it at any point should also be the angle made by the vector $\overrightarrow { E }$ . So $\frac { dy }{ dx } =\frac { 1 }{ 2x }$ That is $dy =\frac{dx}{2x}$ Integrating both the sides. $\int _{ 1 }^{ y }{ 2dy } =\int _{ 1 }^{ x }{ \frac { dx }{ x } } \\ \Rightarrow 2\left( y-1 \right) =\ln { x }$ This will be the equation of the field line passing through $(1,1)$ for the given electric field. Let the particle cross the x-axis at point (a,o).Hence $2\left( 0-1 \right) =\ln { a } \\ \Rightarrow \quad a=\frac { 1 }{ { e }^{ 2 } }$ As electrostatic force is conservative in nature. Using energy conservation $\frac { 1 }{ 2 } m{ v }^{ 2 }=q{ (V }_{ initial }-{ V }_{ final })\\ { v }^{ 2 }=(1+1)-\left( \frac { 1 }{ { e }^{ 4 } } +0 \right) \\ v=\frac { \sqrt { \left( 2{ e }^{ 4 }-1 \right) } }{ { e }^{ 2 } }$ Hence $\alpha=2$ , $\beta=2$ , $\gamma=4$ and $\delta=1$ . So $\alpha$ + $\beta$ + $\gamma$ + $\delta$ = $\boxed{9}$