The tangents drawn from a point P to the ellipse make angles ` theta_(1) and theta_(2)` with the major axis , find the locus of P when
`theta _(1) + theta _(2) ` is constant ( = 2 a)

Answer

To find the locus of the point \( P \) from which tangents are drawn to the ellipse such that the sum of the angles \( \theta_1 + \theta_2 \) is constant (equal to \( 2a \)), we can follow these steps:

### Step 1: Write the equation of the ellipse
The standard equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 2: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse at point \( (x_1, y_1) \) is:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
where \( m \) is the slope of the tangent.

### Step 3: Consider the point \( P \)
Let the coordinates of point \( P \) be \( (α, β) \). For the tangent to pass through \( P \), the coordinates must satisfy the tangent equation:
\[ β = mα + \sqrt{a^2 m^2 + b^2} \]
Rearranging gives:
\[ β - mα = \sqrt{a^2 m^2 + b^2} \]

### Step 4: Square both sides
Squaring both sides results in:
\[ (β - mα)^2 = a^2 m^2 + b^2 \]
Expanding the left side:
\[ β^2 - 2mαβ + m^2α^2 = a^2 m^2 + b^2 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ m^2(α^2 - a^2) - 2mαβ + (β^2 - b^2) = 0 \]
This is a quadratic equation in \( m \).

### Step 6: Use the properties of roots
Let \( m_1 \) and \( m_2 \) be the roots of the quadratic equation. By Vieta's formulas:
- The sum of the slopes \( m_1 + m_2 = \frac{2αβ}{α^2 - a^2} \)
- The product of the slopes \( m_1 m_2 = \frac{β^2 - b^2}{α^2 - a^2} \)

### Step 7: Use the condition \( \theta_1 + \theta_2 = 2a \)
Using the tangent addition formula:
\[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} \]
This gives:
\[ \tan(2a) = \frac{m_1 + m_2}{1 - m_1 m_2} \]

### Step 8: Substitute the values of \( m_1 + m_2 \) and \( m_1 m_2 \)
Substituting the values from Vieta's formulas:
\[ \tan(2a) = \frac{\frac{2αβ}{α^2 - a^2}}{1 - \frac{β^2 - b^2}{α^2 - a^2}} \]

### Step 9: Simplify the equation
Cross-multiplying and simplifying gives:
\[ 2αβ = \tan(2a)(α^2 - β^2 + b^2 - a^2) \]

### Step 10: Write the locus equation
Rearranging gives the locus of point \( P \):
\[ 2xy = \tan(2a)(x^2 - y^2 + b^2 - a^2) \]

### Final Locus Equation
Thus, the locus of point \( P \) is:
\[ 2xy = \tan(2a)(x^2 - y^2 + b^2 - a^2) \]