If the normal to the ellipse `(x^(2))/9+(y^(2))/4=1` which is farthest from its centre be `

Explanation

To find the normal to the ellipse \(\frac{x^2}{9} + \frac{y^2}{4} = 1\) that is farthest from its center, we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse is given in the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where:
- \(a^2 = 9\) implies \(a = 3\)
- \(b^2 = 4\) implies \(b = 2\)

### Step 2: Write the equation of the normal
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