If  two tangents drawn from a point $(\alpha ,\beta )$  lying on the ellipse  $25x^2+4y^2=1$  to the parabola $y^2=4x$  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…………

Equation of tangent to   $y^2=4xisy=mx+\frac{1}{m}$
Satisfy the point  $(\alpha ,\beta )$
Then, the equation of tangent is  $m^2\alpha -\beta m+1=0$ .
It has two roots, $m_1$ and $m_2$  where  $m_1=4m_2$
 $m_1+m_2=\frac{\beta }{\alpha }\Rightarrow m_1{m}_2=\frac{1}{\alpha }$
Put the value of  $m_1$  in both equations.
$\Rightarrow 5m_2=\frac{\beta }{\alpha },4m_2^2=\frac{1}{\alpha }$ 
Solve the above eq’s . Then  $4{\beta }^2=25\alpha$ ------------(i)
Satisfy the points on elipse then, $=25{\alpha }^2+4{\beta }^2=1$  ---------------(ii)
Now, solve equations (i) and (ii).
$25({\alpha }^2+\alpha )=1$ 
Take,  ${(10\alpha +5)}^2+{(16{\beta }^2+50)}^2=25{(2\alpha +1)}^2+2500{(2\alpha +1)}^2$ 
 $=2525(4{\alpha }^2+4\alpha +1)$
 $=2525\times \frac{29}{25}\left\{from\left(iii\right)\right\}=2929.$