The locus of the point of Interection of two tangents to the hyperbola `x^(2)/a^(2)-y^(2)/b^(2)=1` which a make an angle `60^@` with one another is

Options

A. `(x^(2)+y^(2) -a^(2)+b^(2))^(2)=12(a^(2)y^(2)-b^(2)x^(2)+a^(2)b^(2))`
B. `(x^(2)+y^(2) -a^(2)+b^(2))^(2)=4(a^(2)y^(2)-b^(2)x^(2)+a^(2)b^(2))`
C. `3(x^(2)+y^(2) -a^(2)+b^(2))^(2)=4(a^(2)y^(2)-b^(2)x^(2)+a^(2)b^(2))`
D. `x^(2)+y^(2)=a^(2)-b^(2^(2)`

Answer

To find the locus of the point of intersection of two tangents to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) that make an angle of \( 60^\circ \) with one another, we will follow these steps:

### Step 1: Equation of the Tangent
The equation of the tangent to the hyperbola at a point \( (x_1, y_1) \) can be expressed in slope form as:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
where \( m \) is the slope of the tangent.

### Step 2: Point of Intersection
Let \( K(x_1, y_1) \) be the point of intersection of the two tangents. The point \( K \) must satisfy the tangent equation:
\[ y_1 - mx_1 = \pm \sqrt{a^2 m^2 - b^2} \]

### Step 3: Square Both Sides
Squaring both sides gives:
\[ (y_1 - mx_1)^2 = a^2 m^2 - b^2 \]

### Step 4: Expand the Equation
Expanding the left-hand side:
\[ y_1^2 - 2mx_1y_1 + m^2x_1^2 = a^2 m^2 - b^2 \]

### Step 5: Rearranging Terms
Rearranging the equation, we get:
\[ m^2(x_1^2 - a^2) - 2mx_1y_1 + (y_1^2 + b^2) = 0 \]
This is a quadratic equation in \( m \).

### Step 6: Roots of the Quadratic
Let \( m_1 \) and \( m_2 \) be the slopes of the tangents. According to Vieta's formulas:
- The sum of the roots \( m_1 + m_2 = \frac{2x_1y_1}{x_1^2 - a^2} \)
- The product of the roots \( m_1 m_2 = \frac{y_1^2 + b^2}{x_1^2 - a^2} \)

### Step 7: Angle Between the Tangents
The angle \( \theta \) between the tangents is given by:
\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]
For \( \theta = 60^\circ \), we have:
\[ \tan 60^\circ = \sqrt{3} \]

### Step 8: Substitute Values
Substituting the values into the equation gives:
\[ \sqrt{3} = \frac{m_1 - m_2}{1 + m_1 m_2} \]
We can express \( m_1 - m_2 \) in terms of \( m_1 + m_2 \) and \( m_1 m_2 \).

### Step 9: Square Both Sides Again
Squaring both sides and simplifying leads to:
\[ 3(1 + m_1 m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2 \]

### Step 10: Substitute for \( m_1 + m_2 \) and \( m_1 m_2 \)
Substituting the expressions for \( m_1 + m_2 \) and \( m_1 m_2 \) into the equation yields:
\[ 3\left(1 + \frac{y_1^2 + b^2}{x_1^2 - a^2}\right)^2 = \left(\frac{2x_1y_1}{x_1^2 - a^2}\right)^2 - 4\left(\frac{y_1^2 + b^2}{x_1^2 - a^2}\right) \]

### Step 11: Simplify and Rearrange
After simplification, we arrive at the locus equation:
\[ 3x_1^2 + y_1^2 - (a^2 + b^2) = 0 \]

### Final Locus Equation
Thus, the locus of the point of intersection of the two tangents is:
\[ 3x^2 + y^2 = a^2 + b^2 \]