The value of a for the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,(\mathrm{a}>\mathrm{b})$ if the extremities of the latusrectum of the ellipse having positive
ordinates lie on the parabola ${\mathrm{x}}^2=-2(\mathrm{y}-2)$ is 

$\left(\pm \mathrm{ae},\frac{{\mathrm{b}}^2}{\mathrm{a}}\right)$ are extremities of the latusrectum having positive ordinates. Then,

$\begin{array}{r}{\mathrm{a}}^2{\mathrm{e}}^2=-2\left(\frac{{\mathrm{b}}^2}{\mathrm{a}}-2\right)----\left(1\right)\\ \mathrm{But} {\mathrm{b}}^2={\mathrm{a}}^2\left(1-{\mathrm{e}}^2\right)----\left(2\right)\end{array}$

Therefore, from (1) and (2), we get

$\begin{array}{l}{\mathrm{a}}^2{\mathrm{e}}^2-2{\mathrm{ae}}^2+2\mathrm{a}-4=0\\ \mathrm{or} {\mathrm{ae}}^2(\mathrm{a}-2)+2(\mathrm{a}-2)=0\\ \therefore \left({\mathrm{ae}}^2+2\right)(\mathrm{a}-2)=0\end{array}$

Hence, a=2.