$\begin{array}{1 1}(A)xyy'' -x(y')^2 - yy' = 0 \\(B)\;xyy'' + x(y')^2 - yy' = 0 \\(C)\;xyy'' + x(y')^2 + yy' = 0 \\(D)\;xyy'' + x(y')^2 +y' = 0 \end{array} $

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- It is given that the ellipse is having foci on y-axis and the centre at the origin
- Hence the equation of the ellipse is $\large\frac{x^2}{b^2} +\frac{ y^2}{a^2 }$$= 1$

Step 1:

Using the information from the tool box we write the equation as

$\large\frac{x^2}{b^2} +\frac{ y^2}{a^2}$$ = 1$--------(1)

Differentiating this on both sides we get,

$\large\frac{2x}{b^2} + \frac{2yy'}{b^2}$$ = 0$--------(2)

Step 2:

Again differentiating this we get,

$\large\frac{1}{b^2} + (\frac{1}{a^2})$$[yy'' +(y')^2] = 0$

$\large\frac{1}{b^2} = - [\frac{1}{a^2}]$$(yy'' + (y')^2 $

Substituting this in equation (2) we get,

$-[\large\frac{x}{a^2}]$$[yy'' + (y')^2] +\large\frac{ yy'}{a^2}$$= 0$

On simplifying we get

$xyy'' + x(y')^2 - yy' = 0$

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