The tangent at any point P on the ellipse meets the tangents at the vertices A & `A^1` of the ellipse `x^2/a^2 + y^2/b^2 = 1` at L and M respectively. Then `AL.A^1M`= (A) `a^2` (B) `b^2` (C) `a^2+b^2` (D) `ab`

Options

A. a + b
B. `a^(2) + b^(2)`
C. `a^(2)`
D. `b^(2)`

Answer

To solve the problem, we need to find the product of the lengths \( AL \) and \( A'M \) where \( L \) and \( M \) are the points where the tangents at the vertices \( A \) and \( A' \) of the ellipse intersect with the tangent at point \( P \) on the ellipse.

### Step-by-Step Solution:

1. **Identify the Ellipse and Vertices**:
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The vertices of the ellipse are \( A(a, 0) \) and \( A'(-a, 0) \).

2. **Parametric Coordinates of Point P**:
A point \( P \) on the ellipse can be represented parametrically as:
\[ P(A \cos \theta, B \sin \theta) \]

3. **Equation of Tangent at Point P**:
The equation of the tangent to the ellipse at point \( P \) is given by:
\[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]

4. **Finding Point L**:
To find point \( L \), we substitute \( x = -a \) into the tangent equation:
\[ \frac{-a}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]
Simplifying gives:
\[ -\frac{1}{\cos \theta} + \frac{y}{b \sin \theta} = 1 \implies \frac{y}{b \sin \theta} = 1 + \frac{1}{\cos \theta} \]
Thus,
\[ y = b \sin \theta \left(1 + \frac{1}{\cos \theta}\right) = b \sin \theta \left(\frac{\cos \theta + 1}{\cos \theta}\right) \]
Therefore, the coordinates of point \( L \) are:
\[ L\left(-a, \frac{b(1 + \cos \theta)}{\cos \theta}\right) \]

5. **Finding Point M**:
To find point \( M \), we substitute \( x = a \) into the tangent equation:
\[ \frac{a}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]
Simplifying gives:
\[ \frac{1}{\cos \theta} + \frac{y}{b \sin \theta} = 1 \implies \frac{y}{b \sin \theta} = 1 - \frac{1}{\cos \theta} \]
Thus,
\[ y = b \sin \theta \left(1 - \frac{1}{\cos \theta}\right) = b \sin \theta \left(\frac{\cos \theta - 1}{\cos \theta}\right) \]
Therefore, the coordinates of point \( M \) are:
\[ M\left(a, \frac{b(\cos \theta - 1)}{\cos \theta}\right) \]

6. **Calculating Distances AL and A'M**:
The distance \( AL \) is the vertical distance between points \( A \) and \( L \):
\[ AL = \left| 0 - \frac{b(1 + \cos \theta)}{\cos \theta} \right| = \frac{b(1 + \cos \theta)}{\cos \theta} \]
The distance \( A'M \) is the vertical distance between points \( A' \) and \( M \):
\[ A'M = \left| 0 - \frac{b(\cos \theta - 1)}{\cos \theta} \right| = \frac{b(1 - \cos \theta)}{\cos \theta} \]

7. **Finding the Product \( AL \cdot A'M \)**:
Now, we compute the product:
\[ AL \cdot A'M = \left(\frac{b(1 + \cos \theta)}{\cos \theta}\right) \cdot \left(\frac{b(1 - \cos \theta)}{\cos \theta}\right) \]
Simplifying gives:
\[ = \frac{b^2(1 - \cos^2 \theta)}{\cos^2 \theta} = \frac{b^2 \sin^2 \theta}{\cos^2 \theta} \]

8. **Final Result**:
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), the product simplifies to:
\[ AL \cdot A'M = b^2 \]

Thus, the final answer is:
\[ \boxed{b^2} \]