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(A) Given the general solution: $a x^2+b y^2=1$ $\ldots$ $(i)$Differentiating with respect to $x$: $2 a x+2 b y \frac{d y}{d x}=0 \Rightarrow a x+b y

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$\frac{1}{\sqrt{2}}$

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(A) Given the general solution: $a x^2+b y^2=1$ $\ldots$ $(i)$Differentiating with respect to $x$: $2 a x+2 b y \frac{d y}{d x}=0 \Rightarrow a x+b y y'=0$.Differentiating again with respect to $x$: $a+b(y')^2+b y y''=0$.From the first derivative, $a = -b y y' / x$. Substituting this into the second derivative equation:$-b y y' / x + b(y')^2 + b y y'' = 0$.Dividing by $b$ (assuming $b 
eq 0$): $-y y' / x + (y')^2 + y y'' = 0$.Multiplying by $x$: $x y y'' + x (y')^2 - y y' = 0$.The order of this differential equation is $\alpha = 2$ and the degree is $\beta = 1$.Substituting these into the ellipse equation $\alpha x^2+\beta y^2=1$, we get $2 x^2+y^2=1$.Rewriting in standard form: $\frac{x^2}{1/2} + \frac{y^2}{1} = 1$.Here, $a^2 = 1/2$ and $b^2 = 1$. Since $b^2 > a^2$, the eccentricity $e$ is given by $e = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

If $\alpha$ and $\beta$ are respectively the order and degree of the differential equation for which $a x^2+b y^2=1$ is the general solution, then the eccentricity of the ellipse $\alpha x^2+\beta y^2=1$ is

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