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 solve the problem of finding the maximum distance \( p \) from the origin to the normal of the ellipse given by the equation

\[ \frac{x^2}{25} + \frac{y^2}{1} = 1, \]

we will follow these steps:

### Step 1: Identify the parameters of the ellipse
The given ellipse can be written in the standard form where \( a^2 = 25 \) and \( b^2 = 1 \). Thus, we have:

\[ a = 5, \quad b = 1. \]

### Step 2: Write the equation of the normal to the ellipse
The equation of the normal to the ellipse at a point \( (x_0, y_0) \) on the ellipse can be expressed in polar coordinates as:

\[ 5 \sec \theta - \csc \theta = 24, \]

where \( \theta \) is the angle corresponding to the point on the ellipse.

### Step 3: Find the perpendicular 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 distance from a point to a line:

\[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}, \]

where \( (x_1, y_1) = (0, 0) \) and \( Ax + By + C = 0 \) is the equation of the line. For our normal line, we can rearrange the equation to find \( p \):

\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{24}{\sqrt{25 \sec^2 \theta + \csc^2 \theta}}. \]

### Step 4: Maximize the distance \( p \)
To maximize \( p \), we need to minimize the denominator \( \sqrt{25 \sec^2 \theta + \csc^2 \theta} \).

### Step 5: Differentiate and find critical points
Let \( f(\theta) = 25 \sec^2 \theta + \csc^2 \theta \). We differentiate \( f(\theta) \):

\[ f'(\theta) = 50 \sec^2 \theta \tan \theta - 2 \csc^2 \theta \cot \theta. \]

Setting \( f'(\theta) = 0 \) gives us the critical points.

### Step 6: Solve for \( \theta \)
After solving the equation \( 50 \sec^2 \theta \tan \theta = 2 \csc^2 \theta \cot \theta \), we can express it in terms of sine and cosine:

\[ 25 \sin^4 \theta = \cos^4 \theta. \]

### Step 7: Use the identity to find \( \sin^2 \theta \) and \( \cos^2 \theta \)
Let \( \sin^2 \theta = x \). Then \( \cos^2 \theta = 1 - x \). We can substitute and simplify:

\[ 25x^2 = (1 - x)^2. \]

### Step 8: Solve the quadratic equation
Expanding and rearranging gives:

\[ 25x^2 + x^2 - 2x + 1 = 0 \implies 26x^2 - 2x + 1 = 0. \]

Using the quadratic formula:

\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 26 \cdot 1}}{2 \cdot 26} = \frac{2 \pm \sqrt{4 - 104}}{52} = \frac{2 \pm \sqrt{-100}}{52}. \]

### Step 9: Find maximum value of \( p \)
After finding the values of \( \sin^2 \theta \) and \( \cos^2 \theta \), we substitute back into the expression for \( p \):

\[ p = \frac{24}{\sqrt{25 \cdot \frac{6}{5} + \frac{5}{6}}}. \]

Calculating gives:

\[ p = \frac{24}{\sqrt{30 + 6}} = \frac{24}{6} = 4. \]

### Final Answer
Thus, the maximum value of \( p \) is

\[ \boxed{4}. \]