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:

We start with the parabola $$y^{2}=x$$ and the point $$(\alpha ,\beta)$$ on it, where $$\beta>0$$.

Because the point lies on the parabola, substituting its coordinates we immediately obtain the relation $$\beta^{2}=\alpha.$$ This simple fact will be used repeatedly.

To write the equation of the tangent at $$(\alpha ,\beta)$$ we first need its slope. Differentiating $$y^{2}=x$$ with respect to $$x$$ gives $$2y\frac{dy}{dx}=1,$$ so the slope at any point is $$\displaystyle \frac{dy}{dx}=\frac{1}{2y}.$$ Hence, at $$(\alpha ,\beta)$$ the slope is $$m=\dfrac{1}{2\beta}.$$

Using the point-slope form of a straight line, $$y-\beta=m\,(x-\alpha),$$ we write the tangent as $$y-\beta=\frac{1}{2\beta}\,(x-\alpha).$$ Multiplying every term by $$2\beta$$ to clear the fraction we get $$2\beta y-2\beta^{2}=x-\alpha.$$

Now we move every term to one side so that the line is in the form $$L(x,y)=0$$: $$x-2\beta y+2\beta^{2}-\alpha=0.$$ But we have already noted that $$\alpha=\beta^{2},$$ therefore $$2\beta^{2}-\alpha=2\beta^{2}-\beta^{2}=\beta^{2}.$$ Substituting this, the tangent finally simplifies to $$x-2\beta y+\beta^{2}=0.$$

This same straight line is also a tangent to the ellipse $$x^{2}+2y^{2}=1.$$ For a line to be tangent to a conic, substituting the line into the conic must produce a quadratic equation in the remaining variable whose discriminant is zero.

From the tangent we express $$x$$ in terms of $$y$$: $$x=2\beta y-\beta^{2}.$$ We now replace $$x$$ in the ellipse:

$$\bigl(2\beta y-\beta^{2}\bigr)^{2}+2y^{2}=1.$$

Expanding the square, $$4\beta^{2}y^{2}-4\beta^{3}y+\beta^{4}+2y^{2}=1.$$

Collecting like powers of $$y$$ we get a quadratic in $$y$$: $$(4\beta^{2}+2)\,y^{2}-4\beta^{3}y+\beta^{4}-1=0.$$

For this quadratic to have exactly one real root (tangency condition), its discriminant must vanish. The general quadratic $$Ay^{2}+By+C=0$$ has discriminant $$\Delta=B^{2}-4AC.$$ Here $$A=4\beta^{2}+2,\;B=-4\beta^{3},\;C=\beta^{4}-1.$$ Setting $$\Delta=0$$,

$$(-4\beta^{3})^{2}-4(4\beta^{2}+2)(\beta^{4}-1)=0.$$

That is $$16\beta^{6}-4(4\beta^{2}+2)(\beta^{4}-1)=0.$$

Dividing every term by 4 simplifies the equation to $$4\beta^{6}-(4\beta^{2}+2)(\beta^{4}-1)=0.$$

Factor the bracket: $$4\beta^{2}+2=2(2\beta^{2}+1),$$ so $$4\beta^{6}-2(2\beta^{2}+1)(\beta^{4}-1)=0.$$

Dividing by 2 once more gives $$2\beta^{6}-(2\beta^{2}+1)(\beta^{4}-1)=0.$$

Next we expand the product: $$(2\beta^{2}+1)(\beta^{4}-1)=2\beta^{2}\beta^{4}+\beta^{4}-2\beta^{2}-1=2\beta^{6}+\beta^{4}-2\beta^{2}-1.$$

Substituting this back,

$$2\beta^{6}-\Bigl(2\beta^{6}+\beta^{4}-2\beta^{2}-1\Bigr)=0.$$

Simplifying the left side, the $$2\beta^{6}$$ terms cancel, leaving $$-\beta^{4}+2\beta^{2}+1=0.$$ Multiplying through by $$-1$$ we obtain $$\beta^{4}-2\beta^{2}-1=0.$$

Let $$t=\beta^{2};$$ because $$\beta>0$$, we know $$t>0.$$ The equation becomes $$t^{2}-2t-1=0.$$

Using the quadratic formula $$t=\dfrac{2\pm\sqrt{(-2)^{2}-4(1)(-1)}}{2}=\dfrac{2\pm\sqrt{4+4}}{2}=\dfrac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}.$$

Since $$1-\sqrt{2}<0$$, it is inadmissible. Therefore $$t=1+\sqrt{2}.$$ But $$t=\beta^{2},$$ so $$\beta^{2}=1+\sqrt{2}.$$

Finally, recalling that $$\alpha=\beta^{2},$$ we conclude $$\alpha=1+\sqrt{2}.$$

Hence, the correct answer is Option C.