To find image of P(t2,2t) $\begin{aligned}
& \frac{x-t^2}{1}=\frac{y-2 t}{1}=\frac{-2\left(t^2+2 t+4\right)}{1^2+1^2}=-(t+1)^2-3 \
& x=t^2-(t+1)^2-3=-2 t-4 \
& y=2 t-(t+1)^2-3=-t^2-4 \
& t=\frac{-x-4}{2} \
& \Rightarrow y+4=-\left(\frac{-x-4}{2}\right)^2 \
& \Rightarrow(y+4)=-\frac{(x+4)^2}{4} \
& \Rightarrow x^2=-4 y \
& \Rightarrow \text { Focus }(-4,-5)
\end{aligned}\text { Also, } y=-5 \text { intersect }\begin{aligned} & \therefore(-4)(-1)=(x+4)^2 \ & 4=(x+4)^2 \ & x+4= \pm 2 \ & x=-2,-6 \ & \Rightarrow d=4 \ & a=\frac{1}{2}\left|\begin{array}{ccc}1 & 0 & 1 \ -2 & -5 & 1 \ -6 & -5 & 1\end{array}\right| \ & =\frac{1}{2}|[1(-5+5)+1(10-30)]| \ & =\frac{1}{2}(20) \ & a=10 \ & \therefore \quad a+d=14\end{aligned}$