Eliminating Parameter 'a' to Find the Differential Equation

The problem asks us to find the differential equation that represents the given relation by eliminating the parameter 'a'. The given relation is \( y^2 - 2ay + x^2 = a^2 \), and we are given that \( \frac{dy}{dx} = p \).

Step-by-Step Elimination Process

Let's start with the given equation:

\[y^2 - 2ay + x^2 = a^2 \quad (1)\]

Our goal is to eliminate the parameter 'a'. A common method for eliminating a parameter is to differentiate the given equation with respect to x, and then use the original equation and the differentiated equation to remove 'a'.

Differentiating with Respect to x

Differentiate both sides of equation (1) with respect to x. Remember that 'a' is a constant and y is a function of x.

\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(2ay) + \frac{d}{dx}(x^2) = \frac{d}{dx}(a^2) \]

Using the chain rule for \(y^2\) and \(2ay\), and knowing that the derivative of a constant (\(a^2\)) is 0:

\[ 2y \frac{dy}{dx} - 2a \frac{dy}{dx} + 2x = 0 \]

Substitute \( \frac{dy}{dx} = p \):

\[ 2yp - 2ap + 2x = 0 \]

Divide the entire equation by 2:

\[ yp - ap + x = 0 \quad (2) \]

Expressing 'a' in terms of x, y, and p

From equation (2), we can easily express 'a' in terms of x, y, and p:

\[ ap = yp + x \]

\[ a = \frac{yp + x}{p} \quad (3) \]

Substituting 'a' back into the Original Equation

Now substitute the expression for 'a' from equation (3) back into the original equation (1):

\[ y^2 - 2\left(\frac{yp + x}{p}\right)y + x^2 = \left(\frac{yp + x}{p}\right)^2 \]

Simplify the terms:

\[ y^2 - \frac{2y(yp + x)}{p} + x^2 = \frac{(yp + x)^2}{p^2} \]

Clearing Denominators and Expanding

Multiply the entire equation by \(p^2\) to eliminate the denominators:

\[ y^2 p^2 - 2y(yp + x)p + x^2 p^2 = (yp + x)^2 \]

Expand the terms:

\[ y^2 p^2 - 2y^2 p^2 - 2xyp + x^2 p^2 = (yp)^2 + 2(yp)x + x^2 \]

\[ y^2 p^2 - 2y^2 p^2 - 2xyp + x^2 p^2 = y^2 p^2 + 2xyp + x^2 \]

Combining Like Terms

Combine the \(y^2 p^2\) terms on the left side:

\[ (1 - 2)y^2 p^2 - 2xyp + x^2 p^2 = y^2 p^2 + 2xyp + x^2 \]

\[ -y^2 p^2 - 2xyp + x^2 p^2 = y^2 p^2 + 2xyp + x^2 \]

Rearranging to Form the Differential Equation

Move all terms to one side (e.g., move the left-side terms to the right side) to match the format of the options:

\[ 0 = y^2 p^2 + 2xyp + x^2 + y^2 p^2 + 2xyp - x^2 p^2 \]

Combine like terms:

\[ 0 = (y^2 p^2 + y^2 p^2) + (2xyp + 2xyp) + x^2 - x^2 p^2 \]

\[ 0 = 2y^2 p^2 + 4xyp + x^2 - x^2 p^2 \]

Rearrange the terms to group by powers of p and constant terms, and match the options format (e.g., \(p^2\) terms first):

\[ x^2 p^2 - 2y^2 p^2 - 4xyp - x^2 = 0 \]

Factor out \(p^2\):

\[ (x^2 - 2y^2) p^2 - 4xyp - x^2 = 0 \]

This is the differential equation obtained by eliminating the parameter 'a'.

Comparing with Options

Let's compare our derived differential equation with the given options:

Our derived equation matches Option 1. Note that \(4xyp\) is the same as \(4pxy\).

Revision Table: Key Concepts

Additional Information: Families of Curves and Differential Equations

An equation involving a parameter often represents a family of curves. For example, \(x^2 + y^2 = r^2\) represents a family of circles centered at the origin with varying radius 'r'. The differential equation of a family of curves is an equation that is satisfied by every member of that family, regardless of the specific value of the parameter.

The order of the resulting differential equation is typically equal to the number of essential arbitrary parameters in the original relation. In this problem, we started with one parameter 'a', and the resulting differential equation is of the first order (involving only the first derivative, p or \( \frac{dy}{dx} \)).

The process of eliminating parameters is a fundamental concept in the study of differential equations, connecting algebraic equations defining curves to their corresponding differential equations.