If the tangent to the parabola $y^2=x$ at a point $(\alpha ,\beta )$ $(\beta >0)$  is also a tangent to the ellipse,$x^2+2y^2=1$,

then $\alpha$ is equal to 

Let the point of contact be $\left({\beta }^2,\beta \right)\text{. }$

$\therefore$ Tangent to the parabola $y^2=x\text{ at }\left({\beta }^2,\beta \right)$ is

$y\beta =\frac{1}{2}\left(x+{\beta }^2\right)\Rightarrow y=\frac{x}{2\beta }+\frac{\beta }{2}\Rightarrow y=\left(\frac{1}{2\beta }\right)x+\frac{\beta }{2}$

Now, comparing it with $y=mx+c,$we get

$m=\frac{1}{2\beta },c=\frac{\beta }{2}$                                                               $\dots$(i)

Given equation of ellipse is $\frac{x^2}{1}+\frac{y^2}{1/2}=1$

Condition for tangency to ellipse is $c^2=a^2{m}^2+b^2$

$\Rightarrow \frac{{\beta }^2}{4}=\frac{1}{4{\beta }^2}+\frac{1}{2}$                                                         $\text{ [Using (i)] }$

$\Rightarrow \frac{{\beta }^2}{4}=\frac{1+2{\beta }^2}{4{\beta }^2}\Rightarrow {\beta }^4-2{\beta }^2-1=0\Rightarrow {\left({\beta }^2-1\right)}^2=2$

$\Rightarrow {\beta }^2-1=\sqrt{2}\Rightarrow {\beta }^2=1+\sqrt{2}\therefore \alpha =1+\sqrt{2}$