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प्रश्न
Form the differential equation corresponding to (x − a)2 + (y − b)2 = r2 by eliminating a and b.
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उत्तर
The equation of the family of curves is \[\left( x - a \right)^2 + \left( y - b \right)^2 = r^2...............(1)\]
where \[a\text{ and }b\] are parameters.
This equation contains two parameters, so we shall get a second order differential equation.
Differentiating equation (1) with respect to x, we get
\[2\left( x - a \right) + 2\left( y - b \right)\frac{dy}{dx} = 0...............(2)\]
Differentiating (2) with respect to x, we get
\[2 + 2 \left( \frac{dy}{dx} \right)^2 + 2\left( y - b \right)\frac{d^2 y}{d x^2} = 0\]
\[ \Rightarrow 1 + \left( \frac{dy}{dx} \right)^2 + \left( y - b \right)\frac{d^2 y}{d x^2} = 0\]
\[ \Rightarrow \left( y - b \right) = - \frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{d x^2}} .................(3)\]
From (2) and (3), we get
\[\left( x - a \right) - \frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{d x^2}}\frac{dy}{dx} = 0\]
\[ \Rightarrow \left( x - a \right) = \frac{\frac{dy}{dx} + \left( \frac{dy}{dx} \right)^3}{\frac{d^2 y}{d x^2}} .................(4)\]
From (1), (3) and (4), we get
\[ \Rightarrow \frac{\left[ \left( \frac{dy}{dx} \right)^2 + 2 \left( \frac{dy}{dx} \right)^4 + \left( \frac{dy}{dx} \right)^6 \right] + \left[ 1 + 2 \left( \frac{dy}{dx} \right)^2 + \left( \frac{dy}{dx} \right)^4 \right]}{\left( \frac{d^2 y}{d x^2} \right)^2} = r^2 \]
\[ \Rightarrow \left( \frac{dy}{dx} \right)^2 + 2 \left( \frac{dy}{dx} \right)^4 + \left( \frac{dy}{dx} \right)^6 + 1 + 2 \left( \frac{dy}{dx} \right)^2 + \left( \frac{dy}{dx} \right)^4 = r^2 \left( \frac{d^2 y}{d x^2} \right)^2 \]
\[ \Rightarrow 1 + 3 \left( \frac{dy}{dx} \right)^2 + 3 \left( \frac{dy}{dx} \right)^4 + \left( \frac{dy}{dx} \right)^6 = r^2 \left( \frac{d^2 y}{d x^2} \right)^2 \]
\[ \Rightarrow \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^3 = r^2 \left( \frac{d^2 y}{d x^2} \right)^2 \]
It is the required differential equation.
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