If two perpendicular tangents are drawn from a point `(alpha,beta)` to the hyperbola `x^(2) - y^(2) = 16`, then the locus of `(alpha, beta)` is

Options

A. a pair of straight line
B. a circle
C. a parabola
D. an ellipse

Answer

To find the locus of the point \((\alpha, \beta)\) from which two perpendicular tangents can be drawn to the hyperbola given by the equation \(x^2 - y^2 = 16\), we will follow these steps:

### Step 1: Understand the equation of the hyperbola
The hyperbola given is \(x^2 - y^2 = 16\). This can be rewritten in standard form as:
\[ \frac{x^2}{16} - \frac{y^2}{16} = 1 \]
This indicates that the hyperbola opens along the x-axis.

### Step 2: Write the equation of the tangent
The general equation of the tangent to the hyperbola \(x^2 - y^2 = 16\) at a point \((x_0, y_0)\) is given by:
\[ \frac{xx_0}{16} - \frac{yy_0}{16} = 1 \]
If we consider a point \((\alpha, \beta)\) from which the tangents are drawn, the equation of the tangents can be expressed in terms of the slope \(m\):
\[ y = mx \pm \sqrt{16m^2 - 16} \]

### Step 3: Condition for perpendicular tangents
For the tangents to be perpendicular, the product of their slopes must equal \(-1\). If the slopes of the tangents are \(m_1\) and \(m_2\), then:
\[ m_1 \cdot m_2 = -1 \]
From the tangent equation, we can express the slopes in terms of the point \((\alpha, \beta)\) and the hyperbola.

### Step 4: Substitute and simplify
Using the tangent equation, we can derive the condition for the slopes:
\[ m_1 = \frac{\beta - \sqrt{16m_1^2 - 16}}{\alpha} \]
\[ m_2 = \frac{\beta + \sqrt{16m_2^2 - 16}}{\alpha} \]
Setting \(m_1 \cdot m_2 = -1\) and substituting the expressions will lead to a condition involving \(\alpha\) and \(\beta\).

### Step 5: Derive the locus equation
After substituting and simplifying, we arrive at the equation:
\[ \beta^2 = 16 + \alpha^2 \]
This represents a circle with center at the origin and radius 4.

### Final Locus
Thus, the locus of the point \((\alpha, \beta)\) from which two perpendicular tangents can be drawn to the hyperbola \(x^2 - y^2 = 16\) is given by the equation of a circle:
\[ \alpha^2 + \beta^2 = 16 \]