If an ellipse with its axes as coordinate axe | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since it passes through $(2,2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1

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$64$

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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since it passes through $(2,2)$,we have $\frac{4}{a^2} + \frac{4}{b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4}$ ... $(i)$. Since it passes through $(3,1)$,we have $\frac{9}{a^2} + \frac{1}{b^2} = 1$ ... $(ii)$. Subtracting $(i)$ from $(ii)$,we get $\frac{8}{a^2} = 1 - \frac{1}{4} = \frac{3}{4}$. Thus,$a^2 = \frac{32}{3}$,which implies $3a^2 = 32$. Substituting $a^2$ into $(ii)$,we get $\frac{9}{32/3} + \frac{1}{b^2} = 1$,so $\frac{27}{32} + \frac{1}{b^2} = 1$. This gives $\frac{1}{b^2} = 1 - \frac{27}{32} = \frac{5}{32}$,so $b^2 = \frac{32}{5}$,which implies $5b^2 = 32$. Therefore,$3a^2 + 5b^2 = 32 + 32 = 64$.

If an ellipse with its axes as coordinate axes,$2a$ and $2b$ as the lengths of its major and minor axes respectively,passes through the points $(2,2)$ and $(3,1)$,then $3a^2+5b^2=$

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