The differential equation of the family of curves,` y^2 = 4a(x+b)(x+b),a,b,in R`, has order and degree respectively equal to :

Options

A. `1,2`
B. `2,1`
C. `2,2`
D. `1,1`

Explanation

To find the order and degree of the differential equation of the family of curves given by the equation \( y^2 = 4a(x+b)(x+b) \), we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ y^2 = 4a(x+b)(x+b) \]
We will differentiate both sides with respect to \( x \).

### Step 2: Apply implicit differentiation
Using implicit differentiation, we differentiate the left side and the right side:
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4a(x+b)(x+b)) \]
Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 4a \cdot \frac{d}{dx}((x+b)(x+b)) \]
Now, we differentiate the right side using the product rule:
\[ \frac{d}{dx}((x+b)(x+b)) = 2(x+b) \]
Thus, we have:
\[ 2y \frac{dy}{dx} = 4a \cdot 2(x+b) \]
This simplifies to:
\[ 2y \frac{dy}{dx} = 8a(x+b) \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Now, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{8a(x+b)}{2y} = \frac{4a(x+b)}{y} \]

### Step 4: Identify the order and degree
In the expression \(\frac{dy}{dx} = \frac{4a(x+b)}{y}\), we see that:
- The highest derivative present is \(\frac{dy}{dx}\), which is of order 1.
- The equation is linear in \(\frac{dy}{dx}\) since it does not involve any powers greater than 1 or any products of derivatives.

Thus, the order of the differential equation is 1 and the degree is also 1.

### Final Answer
The order and degree of the differential equation are:
- **Order**: 1
- **Degree**: 1