If two perpendicular tangents can be drawn from the point `(alpha,beta)` to the hyperbola `x^(2)-y^(2)=a^(2)` then `(alpha,beta)` lies on

Options

A. `y=pmx`
B. `x^(2)+y^(2)=a^(2)`
C. `x^(2)+y^(2)=2a^(2)`
D. `y^(2)=4ax`

Answer

To solve the problem, we need to determine the condition under which two perpendicular tangents can be drawn from the point \((\alpha, \beta)\) to the hyperbola defined by the equation \(x^2 - y^2 = a^2\).

### Step-by-Step Solution:

1. **Understanding the Hyperbola**:
The hyperbola given is \(x^2 - y^2 = a^2\). The standard form of this hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1\).

2. **Equation of Tangent to the Hyperbola**:
The equation of the tangent to the hyperbola at a point can be expressed as:
\[ y = mx \pm \sqrt{a^2 m^2 - a^2} \]
where \(m\) is the slope of the tangent.

3. **Condition for Perpendicular Tangents**:
If two tangents are perpendicular, the product of their slopes \(m_1\) and \(m_2\) must satisfy:
\[ m_1 m_2 = -1 \]

4. **Using the Tangent Equation**:
From the tangent equation, we can express the tangents from the point \((\alpha, \beta)\):
\[ \beta = m\alpha \pm \sqrt{a^2 m^2 - a^2} \]
Rearranging gives:
\[ \beta - m\alpha = \pm \sqrt{a^2 m^2 - a^2} \]

5. **Squaring Both Sides**:
Squaring both sides leads to:
\[ (\beta - m\alpha)^2 = a^2 m^2 - a^2 \]
Expanding this gives:
\[ \beta^2 - 2\beta m\alpha + m^2\alpha^2 = a^2 m^2 - a^2 \]

6. **Rearranging the Equation**:
Rearranging terms leads to:
\[ (1 - a^2)m^2 + 2\beta m\alpha + (\beta^2 + a^2) = 0 \]
This is a quadratic equation in \(m\).

7. **Condition for Real Roots**:
For \(m\) to have real solutions (which represent the slopes of the tangents), the discriminant of this quadratic must be non-negative:
\[ D = (2\beta\alpha)^2 - 4(1 - a^2)(\beta^2 + a^2) \geq 0 \]

8. **Simplifying the Discriminant**:
Simplifying the discriminant gives:
\[ 4\beta^2\alpha^2 - 4(1 - a^2)(\beta^2 + a^2) \geq 0 \]
Dividing through by 4:
\[ \beta^2\alpha^2 - (1 - a^2)(\beta^2 + a^2) \geq 0 \]

9. **Final Condition**:
Rearranging gives:
\[ \beta^2\alpha^2 - \beta^2 + a^2\beta^2 + a^2 - a^4 \geq 0 \]
This leads to the conclusion that:
\[ \beta^2 = \alpha^2 + a^2 \]
Hence, the point \((\alpha, \beta)\) lies on the curve:
\[ \beta^2 = \alpha^2 + a^2 \]

### Final Answer:
The point \((\alpha, \beta)\) lies on the equation:
\[ \beta^2 = \alpha^2 + a^2 \]