If the tangent to the parabola y ^ 2 = x at a point (α, β), (β > 0)...

$⇒ \left(\alpha, \beta\right) = \left({\beta}^{2}, \beta\right)$

Tangent to a parabola ${y}^{2} = 4ax$ at a point \left({x}_{1}, {y}_{1}\right)$is$y{y}_{1} = 2a\left(x + {x}_{1}\right)$</p><p>Therefore, tangent to the parabola${y}^{2} = x$at \left({\beta}^{2}, \beta\right)$ is$y ⋅ \beta = \dfrac{1}{2}\left(x + {\beta}^{2}\right)$

$⇒ y = \dfrac{x}{2\beta} + \dfrac{\beta}{2} . . . \left(i\right)$

Tangent to parabola is also a tangent to ellipse

${x}^{2} + 2{y}^{2} = 1 ⇒ {x}^{2} + \dfrac{{y}^{2}}{\left(\dfrac{1}{2}\right)} = 1$

And, a line $y = mx + c$ is a tangent to the ellipse $\dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ if ${c}^{2} = {a}^{2}{m}^{2} + {b}^{2}$

Using the condition ${c}^{2} = {a}^{2}{m}^{2} + {b}^{2},$ we get

$\dfrac{{\beta}^{2}}{4} = {1}^{2}{\left(\dfrac{1}{2\beta}\right)}^{2} + \dfrac{1}{2}$

$⇒ \dfrac{{\beta}^{2}}{4} = \dfrac{1}{4{\beta}^{2}} + \dfrac{1}{2}$

$⇒ {\beta}^{4} - 2{\beta}^{2} - 1 = 0$

Using Sridharacharya's Formula the roots of a quadratic equation $a{x}^{2} + bx + c = 0, a \ne 0$ are $x = \dfrac{- b \pm \sqrt{{b}^{2} - 4ac}}{2a}$

$⇒ {\beta}^{2} = \dfrac{- \left(- 2\right) \pm \sqrt{{\left(- 2\right)}^{2} - 4 \times 1 \times \left(- 1\right)}}{2 \times 1}$

$⇒ {\beta}^{2} = \dfrac{2 \pm \sqrt{8}}{2}$

$⇒ {\beta}^{2} = 1 + \sqrt{2}$ or ${\beta}^{2} = 1 - \sqrt{2} ($Neglected as ${\beta}^{2}$ is positive$)$

$⇒ {\beta}^{2} = \sqrt{2} + 1$

$⇒ \alpha = \sqrt{2} + 1 .$