If the ellipse $\frac{x^2}{4}+y^2=1$ meets the ellipse $x^2+\frac{y^2}{a^2}=1$in four distinct points and a = b²-5b+7, then b lies in

$\frac{x^2}{4}+\frac{y^2}{1}=1and\frac{x^2}{1}+\frac{y^2}{a^2}=1$

For the two ellipses to intersect at four distinct points $\Rightarrow a>1$

$\begin{array}{l}\Rightarrow b^2-5b+7>1\Rightarrow b^2-5b+6>0\\ \Rightarrow (b-2)(b-3)>0\Rightarrow b\in (-\infty ,2)\cup (3,\infty )\end{array}$