Find the centre of the ellipse, `(x^(2)-ax)/a^(2)+(y^(2)-by)/b^(2)=0`.

Answer

To find the center of the ellipse given by the equation

\[ \frac{x^2 - ax}{a^2} + \frac{y^2 - by}{b^2} = 0, \]

we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the equation to isolate the terms involving \(x\) and \(y\):

\[ x^2 - \frac{ax}{a^2} + y^2 - \frac{by}{b^2} = 0. \]

### Step 2: Completing the Square for \(x\)
We will complete the square for the \(x\) terms. The expression \(x^2 - \frac{ax}{a^2}\) can be rewritten as:

\[ x^2 - \frac{a}{a^2}x = x^2 - \frac{1}{a}x. \]

To complete the square, we take half of the coefficient of \(x\), square it, and add/subtract it:

\[ x^2 - \frac{1}{a}x = \left(x - \frac{1}{2a}\right)^2 - \left(\frac{1}{2a}\right)^2. \]

### Step 3: Completing the Square for \(y\)
Next, we complete the square for the \(y\) terms. The expression \(y^2 - \frac{by}{b^2}\) can be rewritten as:

\[ y^2 - \frac{b}{b^2}y = y^2 - \frac{1}{b}y. \]

Completing the square gives:

\[ y^2 - \frac{1}{b}y = \left(y - \frac{1}{2b}\right)^2 - \left(\frac{1}{2b}\right)^2. \]

### Step 4: Substitute Back into the Equation
Substituting the completed squares back into the equation gives:

\[ \left(x - \frac{1}{2a}\right)^2 - \left(\frac{1}{2a}\right)^2 + \left(y - \frac{1}{2b}\right)^2 - \left(\frac{1}{2b}\right)^2 = 0. \]

### Step 5: Rearranging the Equation
Rearranging this, we have:

\[ \left(x - \frac{1}{2a}\right)^2 + \left(y - \frac{1}{2b}\right)^2 = \left(\frac{1}{2a}\right)^2 + \left(\frac{1}{2b}\right)^2. \]

### Step 6: Identify the Center
From the standard form of the ellipse, we can identify the center of the ellipse. The center \((h, k)\) is given by:

\[ h = \frac{a}{2}, \quad k = \frac{b}{2}. \]

Thus, the center of the ellipse is:

\[ \left(\frac{a}{2}, \frac{b}{2}\right). \]

### Final Answer
The center of the ellipse is \(\left(\frac{a}{2}, \frac{b}{2}\right)\).

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