If the normal to the ellipse `x^2/25+y^2/1=1` is at a distance p from the origin then the maximum value of p is

Answer

To find the maximum distance \( p \) of the normal to the ellipse \( \frac{x^2}{25} + \frac{y^2}{1} = 1 \) from the origin, we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be compared with the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, we have:
- \( a^2 = 25 \) → \( a = 5 \)
- \( b^2 = 1 \) → \( b = 1 \)

### Step 2: Write the equation of the normal
The equation of the normal to the ellipse at a point \( (a \cos \theta, b \sin \theta) \) is given by:
\[ x = a \cos \theta + \frac{b^2}{a} \sin \theta (y - b \sin \theta) \]
Substituting \( a = 5 \) and \( b = 1 \), the equation becomes:
\[ x = 5 \cos \theta + \frac{1}{5} \sin \theta (y - \sin \theta) \]

### Step 3: Find the distance from the origin to the normal
The distance \( p \) from the origin to the line represented by the normal can be calculated using the formula for the perpendicular distance from a point to a line. The normal can be rearranged into the standard line form \( Ax + By + C = 0 \).

The distance \( p \) is given by:
\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \]
Where \( A \), \( B \), and \( C \) are coefficients from the line equation.

### Step 4: Substitute and simplify
From the normal equation, we can derive \( A \), \( B \), and \( C \). After simplification, we find:
\[ p = \frac{b^2 - a^2}{\sqrt{a^2 \sec^2 \theta + b^2 \csc^2 \theta}} \]
Substituting \( a^2 = 25 \) and \( b^2 = 1 \):
\[ p = \frac{1 - 25}{\sqrt{25 \sec^2 \theta + \csc^2 \theta}} = \frac{-24}{\sqrt{25 \sec^2 \theta + \csc^2 \theta}} \]
Since \( p \) must be positive, we take the absolute value:
\[ p = \frac{24}{\sqrt{25 \sec^2 \theta + \csc^2 \theta}} \]

### Step 5: Minimize the denominator
To maximize \( p \), we need to minimize the denominator \( \sqrt{25 \sec^2 \theta + \csc^2 \theta} \).

Let \( t = 25 \tan^2 \theta + \cot^2 \theta \).

Using the AM-GM inequality:
\[ \frac{25 \tan^2 \theta + \cot^2 \theta}{2} \geq \sqrt{25 \tan^2 \theta \cdot \cot^2 \theta} \]
This gives:
\[ 25 \tan^2 \theta + \cot^2 \theta \geq 10 \]

### Step 6: Substitute back to find maximum \( p \)
Substituting the minimum value of \( t \):
\[ p = \frac{24}{\sqrt{10 + 26}} = \frac{24}{\sqrt{36}} = \frac{24}{6} = 4 \]

### Final Answer
Thus, the maximum value of \( p \) is:
\[ \boxed{4} \]