The differential equation of an ellipse whose | vedclass

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Question

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given that the major axis is twice the minor axis,we have $2a = 2(2b)$

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$x+4 y \frac{dy}{dx}=0$

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Given that the major axis is twice the minor axis,we have $2a = 2(2b)$,which implies $a = 2b$. Substituting $a = 2b$ into the equation,we get $\frac{x^2}{(2b)^2} + \frac{y^2}{b^2} = 1$. This simplifies to $\frac{x^2}{4b^2} + \frac{y^2}{b^2} = 1$,or $x^2 + 4y^2 = 4b^2$. Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 + 4y^2) = \frac{d}{dx}(4b^2)$. This gives $2x + 8y \frac{dy}{dx} = 0$. Dividing by $2$,we obtain $x + 4y \frac{dy}{dx} = 0$.

The differential equation of an ellipse whose major axis is twice its minor axis,is

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