Question

Area of the ellipse $(2x+3y-5{)}^2+4(-3x+2y+1{)}^2=52$ is equal to

Answer 1

Answer: 6.28

$\frac{{\left(\frac{2x+3y-5}{\sqrt{13}}\right)}^2}{4}+\frac{{\left(\frac{-3x+2y+1}{\sqrt{13}}\right)}^2}{1}=1$
which is equivalent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a=2,b=1$
$\therefore$ Area of ellipse $=\pi \cdot 2\cdot 1=2\pi$