Tangents are drawn to ellipse `(x^2)/(a^2) + (y^2)/(b^2) = 1` at points `P(theta_1)` and `Q (theta_2)` then the point of intersection of these tangents is

Options

A. `(("acos"(theta_1 + theta_2)/(2))/("cos"(theta_1 - theta_2)/2), ("b sin" (theta_1 + theta_2)/(2))/("cos"(theta_1 - theta_2)/2))`
B. `(("acos"(theta_1 - theta_2)/(2))/("cos"(theta_1 + theta_2)/2), ("b sin" (theta_1 - theta_2)/(2))/("cos"(theta_1 + theta_2)/2))`
C. `(("a sin"(theta_1 + theta_2)/(2))/("sin"(theta_1 - theta_2)/2), ("b cos" (theta_1 + theta_2)/(2))/("sin"(theta_1 - theta_2)/2))`
D. none

Answer

To find the point of intersection of the tangents drawn to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at points \(P(\theta_1)\) and \(Q(\theta_2)\), we can follow these steps:

### Step 1: Write the equations of the tangents
The equations of the tangents at points \(P(\theta_1)\) and \(Q(\theta_2)\) on the ellipse are given by:

1. For point \(P(\theta_1)\):
\[ \frac{x}{a \cos \theta_1} + \frac{y}{b \sin \theta_1} = 1 \]

2. For point \(Q(\theta_2)\):
\[ \frac{x}{a \cos \theta_2} + \frac{y}{b \sin \theta_2} = 1 \]

### Step 2: Set up the equations
We can rewrite these equations as:

1. \(b \sin \theta_1 x + a \cos \theta_1 y = ab\) (Equation 1)
2. \(b \sin \theta_2 x + a \cos \theta_2 y = ab\) (Equation 2)

### Step 3: Eliminate \(y\)
To find the intersection point, we can eliminate \(y\) by manipulating these equations. We can multiply Equation 1 by \(\sin \theta_2\) and Equation 2 by \(\sin \theta_1\):

1. \(b \sin \theta_1 \sin \theta_2 x + a \cos \theta_1 \sin \theta_2 y = ab \sin \theta_2\)
2. \(b \sin \theta_2 \sin \theta_1 x + a \cos \theta_2 \sin \theta_1 y = ab \sin \theta_1\)

Now, subtract the second equation from the first:

\[ (b \sin \theta_1 \sin \theta_2 - b \sin \theta_2 \sin \theta_1)x + (a \cos \theta_1 \sin \theta_2 - a \cos \theta_2 \sin \theta_1)y = ab(\sin \theta_2 - \sin \theta_1) \]

### Step 4: Simplify the equation
The left-hand side simplifies to:

\[ 0x + (a \cos \theta_1 \sin \theta_2 - a \cos \theta_2 \sin \theta_1)y = ab(\sin \theta_2 - \sin \theta_1) \]

This means we can express \(y\) in terms of \(x\):

\[ y = \frac{ab(\sin \theta_2 - \sin \theta_1)}{a(\cos \theta_1 \sin \theta_2 - \cos \theta_2 \sin \theta_1)} \]

### Step 5: Find \(x\) and \(y\) coordinates
Now, we can find the \(x\) coordinate by substituting back into either of the tangent equations. For example, substituting into Equation 1:

\[ x = \frac{ab(\sin \theta_2 - \sin \theta_1)}{b \sin \theta_1 \cos \theta_2 - b \sin \theta_2 \cos \theta_1} \]

### Step 6: Final coordinates of intersection
Thus, the point of intersection of the tangents is given by:

\[ \left( \frac{a(\sin \theta_1 + \sin \theta_2)}{\sin(\theta_2 - \theta_1)}, \frac{b(\sin \theta_1 + \sin \theta_2)}{\sin(\theta_2 - \theta_1)} \right) \]