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Question
Form the differential equation of the family of curves represented by the equation (a being the parameter):
(2x + a)2 + y2 = a2
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Solution
The equation of the family of curves is \[\left( 2x + a \right)^2 + y^2 = a^2\] ...(1)
where a is a parameter.
As this equation has only one parameter, we shall get a differential equation of first order.
Differentiating (1) with respect to x, we get
\[2\left( 2x + a \right) \times 2 + 2y\frac{dy}{dx} = 0\] ...(2)
Now, from (1), we get
\[4 x^2 + 4ax + a^2 + y^2 = a^2 \]
\[ \Rightarrow 4ax = - y^2 - 4 x^2 \]
\[ \Rightarrow a = - \frac{\left( 4 x^2 + y^2 \right)}{4x}\]
Putting the value of a in (2), we get
\[4\left( 2x - \frac{4 x^2 + y^2}{4x} \right) + 2y\frac{dy}{dx} = 0\]
\[ \Rightarrow 4\left( \frac{8 x^2 - 4 x^2 - y^2}{4x} \right) + 2y\frac{dy}{dx} = 0\]
\[ \Rightarrow 4 x^2 - y^2 + 2xy\frac{dy}{dx} = 0\]
\[ \Rightarrow y^2 - 4 x^2 - 2xy\frac{dy}{dx} = 0\]
It is the required differential equation.
Now, from (1), we get
\[4 x^2 + 4ax + a^2 + y^2 = a^2 \]
\[ \Rightarrow 4ax = - y^2 - 4 x^2 \]
\[ \Rightarrow a = - \frac{\left( 4 x^2 + y^2 \right)}{4x}\]
Putting the value of a in (2), we get
\[4\left( 2x - \frac{4 x^2 + y^2}{4x} \right) + 2y\frac{dy}{dx} = 0\]
\[ \Rightarrow 4\left( \frac{8 x^2 - 4 x^2 - y^2}{4x} \right) + 2y\frac{dy}{dx} = 0\]
\[ \Rightarrow 4 x^2 - y^2 + 2xy\frac{dy}{dx} = 0\]
\[ \Rightarrow y^2 - 4 x^2 - 2xy\frac{dy}{dx} = 0\]
It is the required differential equation.
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