The locus of the point of intersection of two tangents to the hyperbola ` x^(2)//a^(2) -y^(2)//b^(2) = 1` which make an angle ` 90^(@)` with one another is

Options

A. `x^(2)+y^(2)=a^(2)+b^(2)`
B. `x^(2)+y^(2)=a^(2)-b^(2)`
C. `x^(2)-y^(2)=a^(2)-b^(2)`
D. `x^2-y^(2)=a^(2)+b^(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 \( 90^\circ \) with one another, we can follow these steps:

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

### Step 2: Rearranging the Tangent Equation
Rearranging the tangent equation gives:
\[ y - mx = \pm \sqrt{a^2 m^2 - b^2} \]
Squaring both sides results in:
\[ (y - mx)^2 = a^2 m^2 - b^2 \]

### Step 3: Expand the Squared Equation
Expanding the left side:
\[ y^2 - 2mxy + m^2x^2 = a^2 m^2 - b^2 \]

### Step 4: Form a Quadratic in \( m \)
Rearranging gives us a quadratic equation in \( m \):
\[ m^2(x^2 - a^2) - 2kxm + (y^2 + b^2) = 0 \]

### Step 5: Use the Condition for Perpendicular Tangents
For two tangents to be perpendicular, the product of their slopes must equal \(-1\). If \( m_1 \) and \( m_2 \) are the slopes, then:
\[ m_1 m_2 = -1 \]
From the quadratic equation, the product of the roots \( m_1 m_2 \) is given by \( \frac{c}{a} \):
\[ m_1 m_2 = \frac{y^2 + b^2}{x^2 - a^2} = -1 \]

### Step 6: Set Up the Equation
Setting the product of the slopes equal to \(-1\):
\[ \frac{y^2 + b^2}{x^2 - a^2} = -1 \]

### Step 7: Simplify the Equation
Cross-multiplying gives:
\[ y^2 + b^2 = - (x^2 - a^2) \]
Rearranging leads to:
\[ x^2 + y^2 = a^2 - b^2 \]

### Step 8: Final Locus Equation
Thus, the locus of the point of intersection of the tangents is:
\[ x^2 + y^2 = a^2 - b^2 \]

### Conclusion
The final answer is that the locus is a circle with radius \( \sqrt{a^2 - b^2} \).