The maximum value of `5lambda` for which four normals can be drawn to ellipse `x^(2)/25+y^(2)/16=1` through a point (`lambda`,0) is

Answer

To find the maximum value of \( 5\lambda \) for which four normals can be drawn to the ellipse given by the equation

\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]

through the point \((\lambda, 0)\), we can follow these steps:

### Step 1: Identify the parameters of the ellipse
The equation of the ellipse can be compared to the standard form

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a^2 = 25 \) and \( b^2 = 16 \). Thus, we have:

- \( a = 5 \)
- \( b = 4 \)

### Step 2: Write the equation of the normal to the ellipse
The equation of the normal to the ellipse at a point \((x_1, y_1)\) on the ellipse is given by:

\[ \frac{a^2}{x_1^2}(x - x_1) + \frac{b^2}{y_1^2}(y - y_1) = 0 \]

Substituting \( a^2 = 25 \) and \( b^2 = 16 \), we have:

\[ \frac{25}{x_1^2}(x - x_1) + \frac{16}{y_1^2}(y - y_1) = 0 \]

### Step 3: Substitute the point \((\lambda, 0)\)
Since the normal passes through the point \((\lambda, 0)\), we substitute \( x = \lambda \) and \( y = 0 \):

\[ \frac{25}{x_1^2}(\lambda - x_1) + \frac{16}{y_1^2}(0 - y_1) = 0 \]

This simplifies to:

\[ \frac{25}{x_1^2}(\lambda - x_1) - \frac{16}{y_1^2}y_1 = 0 \]

### Step 4: Rearrange the equation
Rearranging gives:

\[ \frac{25}{x_1^2}(\lambda - x_1) = \frac{16}{y_1^2}y_1 \]

### Step 5: Use the parametric equations of the ellipse
The parametric equations for the ellipse are:

\[ x_1 = 5 \cos \theta, \quad y_1 = 4 \sin \theta \]

Substituting these into the equation gives:

\[ \frac{25}{(5 \cos \theta)^2}(\lambda - 5 \cos \theta) = \frac{16}{(4 \sin \theta)^2}(4 \sin \theta) \]

This simplifies to:

\[ \frac{25}{25 \cos^2 \theta}(\lambda - 5 \cos \theta) = \frac{16}{16 \sin^2 \theta}(4 \sin \theta) \]

Thus:

\[ \frac{\lambda - 5 \cos \theta}{\cos^2 \theta} = \frac{4}{\sin^2 \theta} \]

### Step 6: Solve for \(\lambda\)
Cross-multiplying gives:

\[ (\lambda - 5 \cos \theta) \sin^2 \theta = 4 \cos^2 \theta \]

Expanding this:

\[ \lambda \sin^2 \theta - 5 \cos^2 \theta \sin^2 \theta = 4 \cos^2 \theta \]

Rearranging gives:

\[ \lambda \sin^2 \theta = 5 \cos^2 \theta \sin^2 \theta + 4 \cos^2 \theta \]

### Step 7: Factor out \(\cos^2 \theta\)
Factoring out \(\cos^2 \theta\):

\[ \lambda \sin^2 \theta = \cos^2 \theta (5 \sin^2 \theta + 4) \]

### Step 8: Find the maximum value of \(\lambda\)
To find the maximum value of \( \lambda \), we need to analyze the equation. The maximum value of \( \lambda \) occurs when \( \sin^2 \theta \) and \( \cos^2 \theta \) are at their maximum values.

The maximum value of \( \cos^2 \theta \) is 1, hence:

\[ \lambda = 5 \cdot 1 + 4 = 9 \]

### Step 9: Calculate \( 5\lambda \)
Thus, the maximum value of \( 5\lambda \) is:

\[ 5\lambda = 5 \cdot 9 = 45 \]

### Final Answer
Therefore, the maximum value of \( 5\lambda \) for which four normals can be drawn to the ellipse through the point \((\lambda, 0)\) is

\[ \boxed{45} \]