\hspace{-16}$The Curve $\mathbf{y=(\mid x \mid-1)sgn(x-1)}$ Divides $\mathbf{\frac{9x^2}{64}+\frac{4y^2}{25}=\frac{1}{\pi}}$\\\\\\ in Two parts having Area $\mathbf{A_{1}}$ and $\mathbf{A_{2}},$ Where $\mathbf{(A_{1}>A_{2})}$\\\\\\ Then $\mathbf{\frac{A_{1}}{A_{2}}=}$
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