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 (= d)

Answer

To find the locus of point P from which tangents are drawn to the ellipse, we start with the following steps:

### Step 1: Write the equation of the ellipse
The standard form of the 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 any point can be expressed as:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
where \( m \) is the slope of the tangent.

### Step 3: Set the coordinates of point P
Let the coordinates of point P be \( (α, β) \). Since tangents are drawn from this point to the ellipse, the point must satisfy the tangent equation. Thus, we can write:
\[ β = mα + \sqrt{a^2 m^2 + b^2} \]

### Step 4: Rearranging the tangent equation
Rearranging the equation gives:
\[ β - mα = \sqrt{a^2 m^2 + b^2} \]
Now, squaring both sides results in:
\[ (β - mα)^2 = a^2 m^2 + b^2 \]

### Step 5: Expanding the equation
Expanding the left side:
\[ β^2 - 2mαβ + m^2α^2 = a^2 m^2 + b^2 \]
Rearranging gives:
\[ β^2 + b^2 = m^2(α^2 - a^2) + 2mαβ \]

### Step 6: Form a quadratic in terms of m
This can be rearranged to form a quadratic equation in \( m \):
\[ m^2(α^2 - a^2) + 2mαβ + (b^2 - β^2) = 0 \]

### Step 7: Use the relationship between the roots
Let \( m_1 \) and \( m_2 \) be the slopes of the tangents. From the properties of quadratic equations, we know:
- \( m_1 + m_2 = -\frac{2αβ}{α^2 - a^2} \)
- \( m_1 m_2 = \frac{β^2 - b^2}{α^2 - a^2} \)

### Step 8: Relate the slopes to the given condition
Given that \( \tan θ_1 - \tan θ_2 = d \), we have:
\[ m_1 - m_2 = d \]
Using the identity:
\[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 \]
Substituting the values gives:
\[ d^2 = \left(-\frac{2αβ}{α^2 - a^2}\right)^2 - 4\left(\frac{β^2 - b^2}{α^2 - a^2}\right) \]

### Step 9: Simplify the equation
After simplification, we arrive at:
\[ d^2(α^2 - a^2)^2 = 4(α^2β^2 - β^2a^2 + α^2b^2 - ab^2) \]

### Step 10: Final locus equation
Rearranging gives the locus of point P:
\[ d^2(α^2 - a^2)^2 = 4(β^2a^2 + α^2b^2 - a^2b^2) \]