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If an ellipse with its axes as coordinate axes, $2 \mathrm{a}$ and $2 \mathrm{~b}$ as the lengths of its major and minor axes respectively passes through the points $(2,2)$ and $(3,1)$, then $3 a^2+5 b^2=$
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64
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{4}$ ...(i)
$\Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1$ ...(ii)
Equation (ii) - (i),
$\begin{aligned} & \Rightarrow \frac{8}{a^2}=\frac{3}{4} \Rightarrow a^2=\frac{32}{3} \\ & \Rightarrow \quad 3 a^2=32 \\ & \Rightarrow \quad \frac{1}{b^2}=1-\frac{9}{a^2}=1-\frac{27}{32}=\frac{5}{32} \\ & \Rightarrow \quad b^2=\frac{32}{5} \Rightarrow 5 b^2=32 \\ & \therefore \quad 3 a^2+5 b^2=64 .\end{aligned}$
$\Rightarrow \frac{4}{a^2}+\frac{4}{b^2}=1 \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{4}$ ...(i)
$\Rightarrow \frac{9}{a^2}+\frac{1}{b^2}=1$ ...(ii)
Equation (ii) - (i),
$\begin{aligned} & \Rightarrow \frac{8}{a^2}=\frac{3}{4} \Rightarrow a^2=\frac{32}{3} \\ & \Rightarrow \quad 3 a^2=32 \\ & \Rightarrow \quad \frac{1}{b^2}=1-\frac{9}{a^2}=1-\frac{27}{32}=\frac{5}{32} \\ & \Rightarrow \quad b^2=\frac{32}{5} \Rightarrow 5 b^2=32 \\ & \therefore \quad 3 a^2+5 b^2=64 .\end{aligned}$
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