# 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 : Option 1) $\sqrt{2}-1$ Option 2) $2\sqrt{2}-1$   Option 3) $2\sqrt{2}+1$ Option 4) $\sqrt{2} +1$

$y^{2}=x$

since $(\alpha,\beta)$   lies on parabola

$\\\beta^{2}=\alpha\\\\\:(\alpha,\beta)=(\beta^{2},\beta)\\\\\:tangent\:\:to\:\:parabola\\\\\:y.\beta=\frac{1}{2}\left ( x+\beta^{2} \right )\\\\\:y=\frac{x}{2\beta}+\frac{\beta}{2}$

tangent to parabola is also tangent to ellipse

$\\\frac{x^{2}}{4}=1.\left ( \frac{1}{4\beta} \right )+\frac{1}{2}\\\\\:\beta^{2}=\sqrt{2}+1,\alpha=\sqrt{2}+1$

Option 1)

$\sqrt{2}-1$

Option 2)

$2\sqrt{2}-1$

Option 3)

$2\sqrt{2}+1$

Option 4)

$\sqrt{2} +1$

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