Equation of tangent to the parabola y² = x

At (α,β) is T = 0

$\mathrm{y\beta }=\frac{\mathrm{x}+\mathrm{\alpha }}{2}$

$\Rightarrow \mathrm{y\beta }=\frac{\mathrm{x}+{\mathrm{\beta }}^2}{2}(\because {\mathrm{\beta }}^2=\mathrm{\alpha })$

$\Rightarrow \mathrm{y}=\frac{1}{2\mathrm{\beta }}\mathrm{x}+\frac{\mathrm{\beta }}{2}$

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

This is also a tangent to ellipse x² + 2y² = 1

$\therefore \mathrm{C}=\pm \sqrt{{\mathrm{a}}^2{\mathrm{m}}^2+{\mathrm{b}}^2}$

$\Rightarrow \frac{\mathrm{\beta }}{2}=\pm \sqrt{\frac{1}{4{\mathrm{\beta }}^2}+\frac{1}{2}}$

$\Rightarrow \frac{{\mathrm{\beta }}^2}{4}=\frac{1}{4{\mathrm{\beta }}^2}+\frac{1}{2}$

⇒ β⁴ – 2β² – 1 = 0

⇒ (β² – 1) = 2

$\Rightarrow {\mathrm{\beta }}^2-1=\sqrt{2}$

$\Rightarrow {\mathrm{\beta }}^2=\sqrt{2}+1$

$\mathrm{\alpha }={\mathrm{\beta }}^2=\sqrt{2}+1$