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
` tan theta _(1) + tan theta _(2)` is constant (= c)

Answer

To find the locus of the point P from which tangents are drawn to the ellipse such that the sum of the tangents' slopes (tan θ₁ + tan θ₂) is constant, we can follow these steps:

### Step-by-Step Solution:

1. **Equation of the Ellipse**:
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

2. **Equation of the Tangent**:
The equation of the tangent to the ellipse at a point can be expressed as:
\[ y = mx + \sqrt{a^2m^2 + b^2} \]
where \( m \) is the slope of the tangent.

3. **Point P**:
Let the point P be represented as \( (α, β) \). Since the tangent passes through this point, we substitute \( (α, β) \) into the tangent equation:
\[ β = mα + \sqrt{a^2m^2 + b^2} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ β - mα = \sqrt{a^2m^2 + b^2} \]
Now, squaring both sides:
\[ (β - mα)^2 = a^2m^2 + b^2 \]

5. **Expanding and Rearranging**:
Expanding the left side:
\[ β^2 - 2mαβ + m^2α^2 = a^2m^2 + b^2 \]
Rearranging gives:
\[ (α^2 - a^2)m^2 - 2αβm + (β^2 - b^2) = 0 \]

6. **Using Properties of Quadratic Roots**:
In this quadratic equation in terms of \( m \), let \( m_1 \) and \( m_2 \) be the roots (slopes of the tangents). According to Vieta's formulas:
- The sum of the roots \( m_1 + m_2 = \frac{-(-2αβ)}{(α^2 - a^2)} = \frac{2αβ}{α^2 - a^2} \)
- The product of the roots \( m_1m_2 = \frac{β^2 - b^2}{α^2 - a^2} \)

7. **Given Condition**:
We know from the problem that:
\[ \tan θ_1 + \tan θ_2 = c \]
This implies:
\[ m_1 + m_2 = c \]
Therefore, we can equate:
\[ \frac{2αβ}{α^2 - a^2} = c \]

8. **Final Rearrangement**:
Rearranging gives:
\[ 2αβ = c(α^2 - a^2) \]
This can be rewritten as:
\[ 2xy = cx^2 - ca^2 \]

9. **Locus Equation**:
Thus, the locus of point P is given by:
\[ 2xy = cx^2 - ca^2 \]

### Final Answer:
The locus of point P is:
\[ 2xy = cx^2 - ca^2 \]