Question

Form the differential equation of the family of curves represented by the equation (a being the parameter):
(i) (2x + a)² + y² = a²
(ii) (2x − a)² − y² = a²
(iii) (x − a)² + 2y² = a²

Answer 1

(i) 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
$$\begin{gathered} 4x^2+4ax+a^2+y^2=a^2 \\ \Rightarrow 4ax=-y^2-4x^2 \\ \Rightarrow a=-\frac{\left(4x^2+y^2\right)}{4x} \end{gathered}$$
Putting the value of a in (2), we get
$$\begin{gathered} 4\left(2x-\frac{4x^2+y^2}{4x}\right)+2y\frac{dy}{dx}=0 \\ \Rightarrow 4\left(\frac{8x^2-4x^2-y^2}{4x}\right)+2y\frac{dy}{dx}=0 \\ \Rightarrow 4x^2-y^2+2xy\frac{dy}{dx}=0 \\ \Rightarrow y^2-4x^2-2xy\frac{dy}{dx}=0 \\ \text{It is the required differential equation.} \end{gathered}$$

(ii) The equation of the family of curves is
$$\begin{gathered} {\left(2x-a\right)}^2-y^2=a^2 \\ \Rightarrow 4x^2-4ax+a^2-y^2=a^2 \\ \Rightarrow 4x^2-4ax-y^2=0...\left(1\right) \end{gathered}$$
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
$$\begin{gathered} 8x-4a-2y\frac{dy}{dx}=0 \\ \Rightarrow -y\frac{dy}{dx}+4x=2a...\left(2\right) \end{gathered}$$
Now, from (1), we get
$2a=\frac{4x^2-y^2}{2x}$ ...(3)
From (2) and (3), we get
$$\begin{gathered} -y\frac{dy}{dx}+4x=\frac{4x^2-y^2}{2x} \\ \Rightarrow -2xy\frac{dy}{dx}+8x^2=4x^2-y^2 \\ \Rightarrow -2xy\frac{dy}{dx}+4x^2+y^2=0 \\ \Rightarrow 2xy\frac{dy}{dx}=4x^2+y^2 \\ \text{It is the required differential equation.} \end{gathered}$$

(iii) The equation of the family of curves is
$$\begin{gathered} {\left(x-a\right)}^2+2y^2=a^2 \\ \Rightarrow x^2-2ax+a^2+2y^2=a^2 \\ \Rightarrow x^2-2ax+2y^2=0...\left(1\right) \end{gathered}$$
where $a$ is a parameter.
As this equation has only one arbitrary constant, we shall get a differential equation of first order.
Differentiating (1) with respect to $x$, we get
$2x-2a+4y\frac{dy}{dx}=0$ ...(2)
Now, from (1), we get
$2a=\frac{x^2+2y^2}{x}$ ...(3)
From (2) and (3), we get
$$\begin{gathered} 2x-\frac{x^2+2y^2}{x}+4y\frac{dy}{dx}=0 \\ \Rightarrow 2x^2-x^2-2y^2+4xy\frac{dy}{dx}=0 \\ \Rightarrow 4xy\frac{dy}{dx}+x^2-2y^2=0 \\ \Rightarrow 4xy\frac{dy}{dx}=2y^2-x^2 \\ \text{It is the required differential equation.} \end{gathered}$$