If `P(theta),Q(theta+pi/2)` are two points on the ellipse `x^2/a^2+y^2/b^2=1` and α is the angle between normals at P and Q, then

Options

A. `2sqrt(1 - e^(2)) = e sin^(2)2 theta tan alpha`
B. `2sqrt(1 - e^(2)) = e sin^(2) theta tan 2alpha`
C. `sqrt(1 - e^(2)) = 2e^(2) sin^(2) 2theta tan alpha`
D. `2sqrt(1 - e^(2)) = e^(2) sin 2theta tan alpha`

Answer

To solve the problem, we need to find the angle between the normals at two points \( P(\theta) \) and \( Q(\theta + \frac{\pi}{2}) \) on the ellipse given by the equation:

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

### Step 1: Parametric Coordinates of Points \( P \) and \( Q \)

The parametric coordinates of point \( P \) at angle \( \theta \) are:

\[ P = (a \cos \theta, b \sin \theta) \]

For point \( Q \) at angle \( \theta + \frac{\pi}{2} \):

\[ Q = (a \cos(\theta + \frac{\pi}{2}), b \sin(\theta + \frac{\pi}{2})) \]

Using trigonometric identities:

\[ Q = (a \cos(\theta + \frac{\pi}{2}) = -a \sin \theta, \quad b \sin(\theta + \frac{\pi}{2}) = b \cos \theta) \]

Thus, the coordinates of \( Q \) are:

\[ Q = (-a \sin \theta, b \cos \theta) \]

### Step 2: Slope of the Tangent at Points \( P \) and \( Q \)

The equation of the tangent to the ellipse at point \( P \) is given by:

\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]

To find the slope of the tangent line at point \( P \):

\[ \text{slope of tangent at } P = -\frac{b \cos \theta}{a \sin \theta} \]

The slope of the normal at point \( P \) is the negative reciprocal of the slope of the tangent:

\[ m_{normal P} = \frac{a \sin \theta}{b \cos \theta} \]

For point \( Q \), the tangent line is:

\[ \frac{x (-\sin \theta)}{a} + \frac{y \cos \theta}{b} = 1 \]

The slope of the tangent line at point \( Q \):

\[ \text{slope of tangent at } Q = -\frac{b \sin \theta}{-a \cos \theta} = \frac{b \sin \theta}{a \cos \theta} \]

The slope of the normal at point \( Q \):

\[ m_{normal Q} = -\frac{a \cos \theta}{b \sin \theta} \]

### Step 3: Finding the Angle Between the Normals

The angle \( \alpha \) between the two normals can be found using the formula:

\[ \tan \alpha = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| \]

Substituting \( m_1 = \frac{a \sin \theta}{b \cos \theta} \) and \( m_2 = -\frac{a \cos \theta}{b \sin \theta} \):

\[ \tan \alpha = \left| \frac{-\frac{a \cos \theta}{b \sin \theta} - \frac{a \sin \theta}{b \cos \theta}}{1 + \left(\frac{a \sin \theta}{b \cos \theta}\right)\left(-\frac{a \cos \theta}{b \sin \theta}\right)} \right| \]

### Step 4: Simplifying the Expression

The numerator simplifies to:

\[ -\frac{a \cos \theta}{b \sin \theta} - \frac{a \sin \theta}{b \cos \theta} = -\frac{a}{b} \left( \frac{\cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right) = -\frac{a}{b \sin \theta \cos \theta} \]

The denominator simplifies to:

\[ 1 - \frac{a^2}{b^2} = \frac{b^2 - a^2}{b^2} \]

Thus, we have:

\[ \tan \alpha = \left| \frac{-\frac{a}{b \sin \theta \cos \theta}}{\frac{b^2 - a^2}{b^2}} \right| = \frac{a b^2}{b \sin \theta \cos \theta (b^2 - a^2)} \]

### Final Result

The final expression for \( \tan \alpha \) is:

\[ \tan \alpha = \frac{2ab}{(b^2 - a^2) \sin 2\theta} \]