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(D) The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(0,3)$,we have $\fr

Vedclass · Accepted Answer

$xyy' - y^2 + 9 = 0$

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(D) The equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the ellipse passes through $(0,3)$,we have $\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1$,which gives $b^2 = 9$.Thus,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$.Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{9} y' = 0$,which simplifies to $\frac{x}{a^2} + \frac{y y'}{9} = 0$,or $\frac{1}{a^2} = -\frac{y y'}{9x}$.Substituting $\frac{1}{a^2}$ back into the equation $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$,we get $x^2(-\frac{y y'}{9x}) + \frac{y^2}{9} = 1$.Multiplying by $9$,we get $-x y y' + y^2 = 9$,or $x y y' - y^2 + 9 = 0$.

The differential equation representing the family of ellipses having foci either on the $x$-axis or on the $y$-axis,centered at the origin,and passing through the point $(0,3)$ is:

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