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 ^(2) theta _(1) = tan ^(2) theta _(2) ` is constant `( = lambda)`

Answer

To find the locus of the point \( P \) from which tangents are drawn to the ellipse, we will follow these 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 a point \( (x_1, y_1) \) is given by:
\[ y = mx + \sqrt{a^2 m^2 + b^2} \]
where \( m \) is the slope of the tangent.

### Step 3: Consider the point \( P(\alpha, \beta) \)
Let \( P \) be the point from which the tangents are drawn to the ellipse. The coordinates of \( P \) are \( (\alpha, \beta) \).

### Step 4: Substitute the point \( P \) into the tangent equation
Since \( P \) lies on the tangent line, we can substitute \( \alpha \) and \( \beta \) into the tangent equation:
\[ \beta = m\alpha + \sqrt{a^2 m^2 + b^2} \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \beta - m\alpha = \sqrt{a^2 m^2 + b^2} \]
Squaring both sides leads to:
\[ (\beta - m\alpha)^2 = a^2 m^2 + b^2 \]

### Step 6: Expand and simplify
Expanding the left-hand side:
\[ \beta^2 - 2m\alpha\beta + m^2\alpha^2 = a^2 m^2 + b^2 \]
Rearranging gives:
\[ \beta^2 + (m^2\alpha^2 - a^2 m^2) - 2m\alpha\beta - b^2 = 0 \]

### Step 7: Consider the condition \( \tan^2 \theta_1 = \tan^2 \theta_2 = \lambda \)
This implies that the slopes \( m_1 \) and \( m_2 \) of the tangents satisfy:
\[ m_1^2 = m_2^2 = \lambda \]
Thus, we can denote \( m_1 = \sqrt{\lambda} \) and \( m_2 = -\sqrt{\lambda} \).

### Step 8: Use the properties of the roots
Using Vieta's formulas, we know:
- \( m_1 + m_2 = -\frac{b}{a} \)
- \( m_1 m_2 = \frac{b^2 - \lambda a^2}{a^2} \)

### Step 9: Substitute \( m_1 \) and \( m_2 \) into the equations
Substituting \( m_1 \) and \( m_2 \) gives:
\[ 2\sqrt{\lambda} = -\frac{b}{a} \]
and
\[ \lambda = \frac{b^2 - \lambda a^2}{a^2} \]

### Step 10: Find the locus of point \( P \)
After simplifying the equations, we arrive at the locus of point \( P \):
\[ \frac{2\alpha^2\beta^2}{\alpha^2 - a^2} = \lambda(\alpha^2 - a^2) + \beta^2 - b^2 \]

### Final Result
The locus of the point \( P(\alpha, \beta) \) is given by:
\[ \frac{2x^2y^2}{x^2 - a^2} = \lambda(x^2 - a^2) + y^2 - b^2 \]