The differential equation of family of curves $x^{2} = 4 b (y + b), b \in R$ is

x (y^{'})^{2} = 2 y y^{'} - x

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x (y^{'})^{2} = y

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x (y^{'})^{2} = x - 2 y y^{'}

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x (y^{'})^{2} = 2 y y^{'} + x

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Solution

The correct option is D $x (y^{'})^{2} = 2 y y^{'} + x$
Given : $x^{2} = 4 b (y + b) \dots (i)$
Differentiate w.r.t. $x,$ we get
$2 x = 4 b y^{'} \dots (i i)$
Using $(i),$ we get
$2 x = \frac{x^{2}}{y + b} y^{'} \dots (i i i)$
Using $(i i), (i i i)$ we get
$2 x = \frac{2 x^{2} (y^{'})^{2}}{2 y y^{'} + x}$
$\Rightarrow x (y^{'})^{2} = 2 y y^{'} + x$

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