Let `A(theta) and B(phi) `be the extrenities of a chord of an emplise. If the slope of AB is equal to the slope of the tangent at a point C(alpha) on the ellipse, then alpha is equal to

Options

A. `(theta+phi)/2`
B. `(theta-phi)/2`
C. `(theta+phi)/2+pi`
D. `(theta+phi)/2-pi`

Answer

To solve the problem step by step, we start by analyzing the given information about the ellipse and the points on it.

### Step 1: Define the ellipse and points A, B, and C
The equation of the ellipse is given by:

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

The points A and B are the extremities of a chord of the ellipse, defined by the angles \(\theta\) and \(\phi\). Therefore, their coordinates are:

- Point A: \(A(\theta) = (a \cos \theta, b \sin \theta)\)
- Point B: \(B(\phi) = (a \cos \phi, b \sin \phi)\)

Point C, which is also on the ellipse, is defined by the angle \(\alpha\):

- Point C: \(C(\alpha) = (a \cos \alpha, b \sin \alpha)\)

### Step 2: Find the slope of the chord AB
The slope \(m_{AB}\) of the chord AB can be calculated using the formula for the slope between two points:

\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} \]

Substituting the coordinates of points A and B:

\[ m_{AB} = \frac{b \sin \phi - b \sin \theta}{a \cos \phi - a \cos \theta} \]

This simplifies to:

\[ m_{AB} = \frac{b (\sin \phi - \sin \theta)}{a (\cos \phi - \cos \theta)} \]

### Step 3: Use the sine and cosine difference formulas
Using the sine and cosine difference identities, we can express the slope in a more manageable form:

\[ m_{AB} = \frac{b}{a} \cdot \frac{\sin(\phi - \theta)}{\cos(\phi - \theta)} \]

Thus, we have:

\[ m_{AB} = \frac{b}{a} \tan\left(\frac{\phi - \theta}{2}\right) \]

### Step 4: Find the slope of the tangent at point C
To find the slope of the tangent at point C, we differentiate the equation of the ellipse implicitly. The derivative \(y'\) (slope of the tangent) is given by:

\[ \frac{d}{dx}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2}\right) = 0 \]

Differentiating, we get:

\[ \frac{2x}{a^2} + \frac{2y y'}{b^2} = 0 \]

Solving for \(y'\):

\[ y' = -\frac{b^2 x}{a^2 y} \]

Substituting \(x = a \cos \alpha\) and \(y = b \sin \alpha\):

\[ y' = -\frac{b^2 (a \cos \alpha)}{a^2 (b \sin \alpha)} = -\frac{b}{a} \cot \alpha \]

### Step 5: Set the slopes equal
Since the slope of the chord AB is equal to the slope of the tangent at point C, we have:

\[ \frac{b}{a} \tan\left(\frac{\phi - \theta}{2}\right) = -\frac{b}{a} \cot \alpha \]

This simplifies to:

\[ \tan\left(\frac{\phi - \theta}{2}\right) = -\cot \alpha \]

### Step 6: Use the cotangent identity
Using the identity \(\cot \alpha = -\tan\left(\frac{\pi}{2} + \alpha\right)\), we can rewrite the equation:

\[ \tan\left(\frac{\phi - \theta}{2}\right) = \tan\left(\frac{\pi}{2} + \alpha\right) \]

### Step 7: Solve for \(\alpha\)
From the tangent identity, we know that if \(\tan A = \tan B\), then:

\[ A = B + n\pi \quad \text{for } n \in \mathbb{Z} \]

Thus, we have:

\[ \frac{\phi - \theta}{2} = \frac{\pi}{2} + \alpha + n\pi \]

Rearranging gives:

\[ \alpha = \frac{\phi - \theta}{2} - \frac{\pi}{2} + n\pi \]

### Conclusion
The general solution for \(\alpha\) is:

\[ \alpha = n\pi + \frac{\phi + \theta}{2} \]