Locus of the point of intersection of the tangents at the points with eccentric angle `theta` and `(pi)/(2)+theta` is

Options

A. `x^(2)+y^(2)=a^(2)`
B. `x^(2)+y^(2)=b^(2)`
C. `(x^(2))/(a^(2))+(y^(2))/(b^(2))=2`
D. `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`

Answer

To find the locus of the point of intersection of the tangents at the points with eccentric angles \( \theta \) and \( \frac{\pi}{2} + \theta \) on the ellipse, we can follow these steps:

### Step 1: Identify the Points on the Ellipse
The points on the ellipse corresponding to the eccentric angles \( \theta \) and \( \frac{\pi}{2} + \theta \) can be represented as:
- Point \( P \) at \( (a \cos \theta, b \sin \theta) \)
- Point \( Q \) at \( (a \cos(\frac{\pi}{2} + \theta), b \sin(\frac{\pi}{2} + \theta)) \)

Using trigonometric identities, we know:
- \( \cos(\frac{\pi}{2} + \theta) = -\sin \theta \)
- \( \sin(\frac{\pi}{2} + \theta) = \cos \theta \)

Thus, the coordinates of point \( Q \) become:
- Point \( Q \) at \( (-a \sin \theta, b \cos \theta) \)

### Step 2: Write the Equation of the Tangent at Point \( P \)
The equation of the tangent to the ellipse at point \( P \) is given by:
\[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]
Multiplying through by \( a \cos \theta \cdot b \sin \theta \) gives:
\[ b \sin \theta \cdot x + a \cos \theta \cdot y = a b \]
This can be rearranged to:
\[ b \sin \theta \cdot x + a \cos \theta \cdot y - ab = 0 \tag{1} \]

### Step 3: Write the Equation of the Tangent at Point \( Q \)
The equation of the tangent to the ellipse at point \( Q \) is:
\[ \frac{x}{-a \sin \theta} + \frac{y}{b \cos \theta} = 1 \]
Multiplying through by \( -a \sin \theta \cdot b \cos \theta \) gives:
\[ b \cos \theta \cdot x - a \sin \theta \cdot y = -ab \]
This can be rearranged to:
\[ b \cos \theta \cdot x - a \sin \theta \cdot y + ab = 0 \tag{2} \]

### Step 4: Find the Point of Intersection of the Two Tangents
To find the point of intersection of the two tangents, we can solve the equations (1) and (2) simultaneously.

Let the point of intersection be \( (h, k) \). Thus, we have:
1. \( b \sin \theta \cdot h + a \cos \theta \cdot k = ab \)
2. \( b \cos \theta \cdot h - a \sin \theta \cdot k = -ab \)

### Step 5: Solve the System of Equations
We can express the equations in terms of \( h \) and \( k \):
- From (1): \( k = \frac{ab - b \sin \theta \cdot h}{a \cos \theta} \)
- Substitute \( k \) into (2):
\[ b \cos \theta \cdot h - a \sin \theta \cdot \left(\frac{ab - b \sin \theta \cdot h}{a \cos \theta}\right) = -ab \]

### Step 6: Simplify and Rearrange
After substituting and simplifying, we will eventually arrive at an equation that relates \( h \) and \( k \).

### Step 7: Final Locus Equation
After simplification, we will find that the locus of the point of intersection is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 2 \]