The locus of a point from which the two tangents to the ellipse are inclined at an angle `alpha`.

Options

A. `tan^2 alpha(x^2+y^2-a^2-b^2)^2=4{x^2b^2+a^2y^2-a^2b^2}`
B. `2tan^2 alpha(x^2+y^2-a^2-b^2)^2=4{x^2b^2+a^2y^2-a^2b^2}`
C. `tan^2 alpha(x^2+y^2-a^2-b^2)^2={x^2b^2+a^2y^2-a^2b^2}`
D. None of these

Answer

To find the locus of a point from which the two tangents to the ellipse are inclined at an angle \( \alpha \), we can follow these steps:

### Step 1: Understand the Equation of the Ellipse
The standard equation of an ellipse is given by:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

where \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.

### Step 2: Set Up the Tangent Equation
The equation of the tangent to the ellipse at a point \( (x_1, y_1) \) on the ellipse can be expressed as:

\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]

### Step 3: Find the Condition for the Angle Between Two Tangents
Let \( m_1 \) and \( m_2 \) be the slopes of the two tangents from a point \( (x_0, y_0) \) to the ellipse. The angle \( \alpha \) between the two tangents can be expressed using the formula:

\[ \tan(\alpha) = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 4: Use the Condition for Tangents
From the point \( (x_0, y_0) \), the slopes of the tangents to the ellipse can be derived from the quadratic equation formed by substituting \( y = mx + c \) into the ellipse equation. The discriminant of this quadratic must be zero for tangents to exist, leading to the condition:

\[ b^2m^2 + 2m\left(\frac{b^2y_0}{a^2} - \frac{a^2x_0}{b^2}\right) + \left(\frac{b^2y_0^2}{a^2} - b^2\right) = 0 \]

### Step 5: Relate the Slopes to the Angle
Using the relationship between the slopes and the angle, we can express:

\[ \tan(\alpha) = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 6: Substitute and Rearrange
Substituting the values of \( m_1 \) and \( m_2 \) in terms of \( x_0 \) and \( y_0 \) into the equation, we can derive a relationship involving \( x_0 \) and \( y_0 \).

### Step 7: Derive the Locus Equation
After substituting and simplifying, we arrive at the locus equation:

\[ \tan^2(\alpha) \left( x^2 + y^2 - (a^2 + b^2) \right)^2 = 4b^2x^2 + 4a^2y^2 - 4(a^2 + b^2) \]

### Conclusion
Thus, the locus of the point from which the two tangents to the ellipse are inclined at an angle \( \alpha \) is given by the above equation.

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