Area formed by the curve 4(x – 2y + 1)² + 9(2x + y + 2)² = 25 if the lines 2x+y+2=0 and x–2y+1=0 are taken as new x and y-axis respectively is k$\mathrm{\pi }$ then k = 

$\begin{array}{l}4(x-2y+1{)}^2+9(2x+y+2{)}^2=25\\ \Rightarrow 4{\left(\frac{x-2y+1}{\sqrt{5}}\right)}^2+9{\left(\frac{2x+y+2}{\sqrt{5}}\right)}^2=5\\ x-2y+1=0,2x+y+2=0\text{ are }\perp \text{ er }\end{array}$
and taken as axes. let p(x,y) be the locus of given equations in new axes. 4X² + 9Y² = 5
$\begin{array}{l}\frac{X^2}{{\displaystyle \frac{5}{4}}}+\frac{Y^2}{{\displaystyle \frac{5}{9}}}=1\text{ it is an ellipse }\\ \Rightarrow a^2=\frac{5}{4}{b}^2=\frac{5}{9}\Rightarrow a=\frac{\sqrt{5}}{2}b=\frac{\sqrt{5}}{3}\end{array}$
Area of ellipse $=\pi ab=\pi \cdot \frac{\sqrt{5}}{2}\cdot \frac{\sqrt{5}}{3}=\frac{5\pi }{6}\text{ squnits }$

$\Rightarrow K=\frac{5}{6}$