In an ellipse `x^(2)/a^(2) + y^(2)/b^(2) = 1` where `a gt b`, the distance of the normal from the centre does not exceed by ?

Answer

To find the maximum distance of the normal from the center of the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a > b \), we will follow these steps:

### Step 1: Parametric Form of the Ellipse
The parametric equations for the ellipse are:
\[ x = a \cos \theta, \quad y = b \sin \theta \]

### Step 2: Equation of the Normal
The equation of the normal to the ellipse at the point \( (x_1, y_1) \) is given by:
\[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \]
Substituting the parametric equations for \( x_1 \) and \( y_1 \):
\[ \frac{a^2 x}{a \cos \theta} - \frac{b^2 y}{b \sin \theta} = a^2 - b^2 \]
This simplifies to:
\[ a x \sec \theta - b y \csc \theta = a^2 - b^2 \]

### Step 3: Distance from the Center to the Normal
The distance \( D \) from the center \( (0, 0) \) to the line \( ax + by + c = 0 \) is given by:
\[ D = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \]
In our case, substituting \( x_1 = 0 \) and \( y_1 = 0 \):
\[ D = \frac{|c|}{\sqrt{a^2 + b^2}} \]
From the normal equation, we can express \( c \) as:
\[ c = -(a^2 - b^2) \]
Thus, the distance becomes:
\[ D = \frac{|-(a^2 - b^2)|}{\sqrt{a^2 + b^2}} = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}} \]

### Step 4: Finding the Maximum Distance
To find the maximum distance, we need to minimize the expression:
\[ D = \frac{a^2 - b^2}{\sqrt{a^2 + b^2}} \]
We can analyze this expression to find its maximum value.

### Step 5: Result
The maximum distance of the normal from the center does not exceed:
\[ |a - b| \]

Thus, the final answer is:
\[ \text{Maximum distance of the normal from the center} = |a - b| \]