If curves ${\text{y=1−ax}}^2 \mathrm{and} \mathrm{y}={\mathrm{x}}^2$  intersect orthogonally then ‘a’ is $\frac{\mathrm{\alpha }}{\mathrm{\beta }}\text{ then }\mathrm{\alpha }+\mathrm{\beta }=$_______________

Point of intersection of $\mathrm{y}=1-{\mathrm{ax}}^2\text{ and }\mathrm{y}={\mathrm{x}}^2$ is 
$\begin{array}{l}1-{\mathrm{ax}}^2={\mathrm{x}}^2\Rightarrow {\mathrm{x}}^2=\frac{1}{1+\mathrm{a}}\\ {\mathrm{m}}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=-2\mathrm{ax}{\mathrm{m}}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=2\mathrm{x}\end{array}$
Curves cut orthogonally
$\begin{array}{l}{\mathrm{m}}_1\times {\mathrm{m}}_2=-1\Rightarrow -2\mathrm{ax}\times 2\mathrm{x}=-1\\ \Rightarrow -4{\mathrm{ax}}^2=-1\\ \Rightarrow 4\mathrm{a}\left(\frac{1}{1+\mathrm{a}}\right)=1\\ \Rightarrow \mathrm{a}=1/3\\ \therefore \mathrm{\alpha }=1\mathrm{\beta }=3\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=1+3=4\end{array}$