If two tangents drawn from a point ($\alpha$, $\beta$) 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$\alpha$ + 5)² + (16$\beta$² + 50)² equals ___________.

$\because(\alpha, \beta)$ lies on the given ellipse, $25 \alpha^{2}+4 \beta^{2}=1\quad\quad...(i)$ <br/><br/> Tangent to the parabola, $y=m x+\frac{1}{m}$ passes through $(\alpha, \beta)$. So, $\alpha m^{2}-\beta m+1=0$ has roots $m_{1}$ and $4 m_{1}$, <br/><br/> $$ m_{1}+4 m_{1}=\frac{\beta}{\alpha} \text { and } m_{1} \cdot 4 m_{1}=\frac{1}{\alpha} $$ <br/><br/> Gives that $4 \beta^{2}=25 \alpha \quad\quad...(ii)$ <br/><br/> from (i) and (ii) <br/><br/> $25\left(\alpha^{2}+\alpha\right)=1\quad\quad...(iii)$ <br/><br/> Now, $(10 \alpha+5)^{2}+\left(16 \beta^{2}+50\right)^{2}$ <br/><br/> $=25(2 \alpha+1)^{2}+2500(2 \alpha+1)^{2}$ <br/><br/> $=2525\left(4 \alpha^{2}+4 \alpha+1\right)$ from equation (iii) <br/><br/> $=2525\left(\frac{4}{25}+1\right)$ <br/><br/> $=2929$