Let point (h, k) lies on x² + y² = 4 and point (2h + 1 , 3k + 2) lies on ellipse having eccentricity e. Then the value of 5 / e² is ______.

The correct answer is 9.

Given that the point $(h,k)$ lies on the circle $x^2+y^2=4$, we have:
$h^2+k^2=4$

Let $(X,Y)$ represent the coordinates of the point lying on the ellipse, such that:
$X=2h+1$
and
$Y=3k+2$

We can express $h$ and $k$ in terms of $X$ and $Y$ respectively:
$h=\frac{X-1}{2}$
$k=\frac{Y-2}{3}$

Substituting these expressions into the equation $h^2+k^2=4$ gives:
${\left(\frac{X-1}{2}\right)}^2+{\left(\frac{Y-2}{3}\right)}^2=4$

Simplifying the denominators:
$\frac{{(X-1)}^2}{4}+\frac{{(Y-2)}^2}{9}=4$

Dividing the entire equation by 4, we obtain the standard equation of the ellipse:
$\frac{{(X-1)}^2}{16}+\frac{{(Y-2)}^2}{36}=1$

Comparing this with the general equation of a shifted ellipse $\frac{{(x-x_0)}^2}{a^2}+\frac{{(y-y_0)}^2}{b^2}=1$, we identify the parameters:
$a^2=16$
$b^2=36$

Since $b^2>a^2$, the ellipse is vertically oriented, and the relation for its eccentricity $e$ is given by:
$a^2=b^2(1-e^2)$

Substituting the values of $a^2$ and $b^2$:
$16=36(1-e^2)$
$1-e^2=\frac{16}{36}$
$1-e^2=\frac{4}{9}$
$e^2=1-\frac{4}{9}=\frac{5}{9}$

Finally, calculating the required value:
$\frac{5}{e^2}=\frac{5}{5/9}=9$