The normal inclined at 45° to the x-axis of the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ drawn and meets the major and minor axis in P and Q respectively. If C is the centre of the ellipse, the area of $\mathrm{\Delta CPQ} \mathrm{is} \frac{{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}^2}{\mathrm{K}\left({\mathrm{a}}^2+{\mathrm{b}}^2\right)}$ square units. Then value of $\frac{{\mathrm{K}}^4}{7}=$…….

Let the equation of the lines be y = x + c ........(1)
If it is a normal of $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$, 
then $\frac{{\mathrm{a}}^2}{1}+\frac{\mathrm{b}}{1}=\frac{{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}^2}{{\mathrm{c}}^2}\Rightarrow {\mathrm{c}}^2=\frac{{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}$
(1) cuts the major axis at P and minor axis at Q 
$\Rightarrow \therefore \mathrm{P}=(-\mathrm{c},0);\mathrm{Q}=(0,\mathrm{c})$
Area of the $\mathrm{\Delta CPQ}=\frac{1}{2}{\mathrm{c}}^2=\frac{1}{2}\frac{{\left({\mathrm{a}}^2-{\mathrm{b}}^2\right)}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}$
$\Rightarrow \therefore \mathrm{K}=2$

$Now \frac{k^4}{7}=\frac{16}{7}=2.28$