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Q.
Area of the ellipse $(2 x+3 y-5)^{2}+4(-3 x+2 y+1)^{2}=52$ is equal to
Conic Sections
Solution:
$\frac{\left(\frac{2 x+3 y-5}{\sqrt{13}}\right)^{2}}{4}+\frac{\left(\frac{-3 x+2 y+1}{\sqrt{13}}\right)^{2}}{1}=1$
which is equivalent to $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $a=2, b=1$
$\therefore $ Area of ellipse $=\pi \cdot 2 \cdot 1=2 \pi$