Find the centre and eccentricity of the ellipse `4(x-2y+1)^(2)+9(2x+y+2)^(2)=5`.

Answer

To find the center and eccentricity of the ellipse given by the equation \(4(x - 2y + 1)^2 + 9(2x + y + 2)^2 = 5\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start by moving \(5\) to the left side of the equation:

\[ 4(x - 2y + 1)^2 + 9(2x + y + 2)^2 - 5 = 0 \]

Now, we can rewrite the equation as:

\[ \frac{(x - 2y + 1)^2}{\frac{5}{4}} + \frac{(2x + y + 2)^2}{\frac{5}{9}} = 1 \]

### Step 2: Identify \(a^2\) and \(b^2\)
From the standard form of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we can identify:

\[ a^2 = \frac{5}{4} \quad \text{and} \quad b^2 = \frac{5}{9} \]

### Step 3: Calculate eccentricity
The eccentricity \(e\) of the ellipse is given by the formula:

\[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Substituting the values of \(a^2\) and \(b^2\):

\[ e = \sqrt{1 - \frac{\frac{5}{9}}{\frac{5}{4}}} \]

This simplifies to:

\[ e = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Step 4: Find the center of the ellipse
To find the center, we need to solve the equations derived from the linear terms in the original equation:

1. \(x - 2y + 1 = 0\)
2. \(2x + y + 2 = 0\)

We can solve these equations simultaneously. From the first equation, we can express \(x\) in terms of \(y\):

\[ x = 2y - 1 \]

Substituting this into the second equation:

\[ 2(2y - 1) + y + 2 = 0 \]

This simplifies to:

\[ 4y - 2 + y + 2 = 0 \implies 5y = 0 \implies y = 0 \]

Now substituting \(y = 0\) back into the equation for \(x\):

\[ x = 2(0) - 1 = -1 \]

Thus, the center of the ellipse is:

\[ (-1, 0) \]

### Final Answer
The center of the ellipse is \((-1, 0)\) and the eccentricity is \(\frac{\sqrt{5}}{3}\).

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