If two tangents drawn from a point (α, ß) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4x are such that the slope of one tangent is four times the other, then the value of $(10\mathrm{\alpha }+5{)}^2+{\left(16{\mathrm{\beta }}^2+50\right)}^2$ equals __________

$\mathrm{\alpha }=\frac{1}{5}\cos \mathrm{\theta },\mathrm{\beta }=\frac{1}{2}\sin \mathrm{\theta }$
Equation of tangent to y² = 4x
$\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{m}}$
It passes through (α,ß)
$\begin{array}{l}\frac{1}{2}\sin \mathrm{\theta }=\mathrm{m}\frac{1}{5}\cos \mathrm{\theta }+\frac{1}{\mathrm{m}}\\ {\mathrm{m}}^2\left(\frac{\cos \mathrm{\theta }}{5}\right)-\mathrm{m}\left(\frac{1}{2}\sin \mathrm{\theta }\right)+1=0\end{array}$
It has two roots m₁ and m₂ where m₁ = 4m₂
$\begin{array}{l}{\mathrm{m}}_1+{\mathrm{m}}_2=\frac{\frac{1}{2}\sin \mathrm{\theta }}{\frac{\cos \mathrm{\theta }}{5}}\\ {\mathrm{m}}_1{\mathrm{m}}_2=\frac{5}{\cos \mathrm{\theta }}\end{array}$
After eliminating m₁ and m₂
$\begin{array}{l}\cos \mathrm{\theta }=\frac{-5\pm \sqrt{29}}{2}\\ \mathrm{\alpha }=\frac{-5\pm \sqrt{29}}{10}\Rightarrow 10\mathrm{\alpha }+5=\pm \sqrt{29}\\ {\mathrm{\beta }}^2=\frac{1}{4}{\sin }^2\mathrm{\theta }\Rightarrow 16{\mathrm{\beta }}^2=-50\pm 10\sqrt{29}\\ (10\mathrm{\alpha }+5{)}^2+{\left(16{\mathrm{\beta }}^2+50\right)}^2=2929\end{array}$