If the ellipse  $\frac{x^2}{4}+\frac{y^2}{1}=1$  meet the ellipse $\frac{x^2}{1}+\frac{y^2}{a^2}=1$   in four distinct points and $a=b^2-10b+25$ , then the value  $b$ does not satisfy

Given two ellipses intersect if second ellipse  $\frac{x^2}{1}+\frac{y^2}{a^2}=1$ is vertical

Now according to condition  $a>1$ 
$b^2-10b+25=a>1$    $\Rightarrow b^2-10b+24>0$
$\Rightarrow (b-4)(b-6)>0$  i.e$b<4\&b>6$

$\Rightarrow b\notin [4,6]$