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In each of the following, $Q .1$ to 5 form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b$.
$y^2=a\left(b^2-x^2\right)$
$y^2=a\left(b^2-x^2\right)$
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$y^2=a\left(b^2-x^2\right)$
Differentiating w.r.t $x ; 2 y y^{\prime}=-2 a x$
Again differentiating, $2\left(y y^{\prime \prime}+y^{\prime} y^{\prime}\right)=-2 a$
$\Rightarrow 2\left(y^{\prime 2}+y y^{\prime \prime}\right)=-2 a$
Dividing (iii) by (ii): $\frac{2\left(\mathrm{y}^{\prime 2}+\mathrm{yy}^{\prime \prime}\right)}{2 \mathrm{yy}^{\prime}}=\frac{-2 \mathrm{a}}{-2 \mathrm{ax}}$ $\Rightarrow x\left(y^{\prime 2}+y y^{\prime \prime}\right)=y^{\prime}$; i.e. the differential equation $x y\left(\frac{d^2 y}{d x^2}\right)+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$
Differentiating w.r.t $x ; 2 y y^{\prime}=-2 a x$
Again differentiating, $2\left(y y^{\prime \prime}+y^{\prime} y^{\prime}\right)=-2 a$
$\Rightarrow 2\left(y^{\prime 2}+y y^{\prime \prime}\right)=-2 a$
Dividing (iii) by (ii): $\frac{2\left(\mathrm{y}^{\prime 2}+\mathrm{yy}^{\prime \prime}\right)}{2 \mathrm{yy}^{\prime}}=\frac{-2 \mathrm{a}}{-2 \mathrm{ax}}$ $\Rightarrow x\left(y^{\prime 2}+y y^{\prime \prime}\right)=y^{\prime}$; i.e. the differential equation $x y\left(\frac{d^2 y}{d x^2}\right)+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$
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