if the chord of contact of tangents from a point P to the hyperbola `x^2/a^2-y^2/b^2=1` subtends a right angle at the centre, then the locus of P is

Options

A. `x^(2)/a^(2) + y^(2)/b^(2) = 1/a^(2) + 1/b^(2)`
B. `x^(2)/a^(4) + y^(2)/b^(4) = (1/a + 1/b)^(2)`
C. `x^(2)/a^(2) + y^(2)/b^(2) = 1/a^(4) + 1/b^(4)`
D. `x^(2)/a^(4) + y^(2)/b^(4) = 1/a^(2) + 1/b^(2)`

Answer

To solve the problem, we need to find the locus of the point \( P(h, k) \) from which tangents are drawn to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) such that the chord of contact subtends a right angle at the center of the hyperbola.

### Step-by-Step Solution:

1. **Identify the Point P**:
Let the point \( P \) be \( (h, k) \).

2. **Equation of the Chord of Contact**:
The equation of the chord of contact of tangents drawn from the point \( P(h, k) \) to the hyperbola is given by:
\[ \frac{hx}{a^2} - \frac{ky}{b^2} = 1 \]

3. **Condition for Right Angle**:
The chord of contact subtends a right angle at the center of the hyperbola. This means the slopes of the tangents from point \( P \) to the hyperbola must satisfy the condition for perpendicularity.

4. **Finding the Slopes**:
The slopes of the tangents to the hyperbola from point \( P(h, k) \) can be derived from the equation of the chord of contact. If we rewrite the equation:
\[ hx - a^2 = ky \cdot \frac{b^2}{b^2} \]
The slopes of the tangents can be found by rearranging the equation into slope-intercept form.

5. **Using the Condition of Perpendicularity**:
For two lines to be perpendicular, the product of their slopes must equal -1. If we denote the slopes of the tangents as \( m_1 \) and \( m_2 \), then:
\[ m_1 \cdot m_2 = -1 \]

6. **Substituting and Simplifying**:
After substituting the slopes into the perpendicularity condition, we can derive a relationship between \( h \) and \( k \). This will yield a quadratic equation in terms of \( h \) and \( k \).

7. **Finding the Locus**:
The locus of point \( P(h, k) \) can be expressed in a standard form. After simplifying the derived equation, we can express it as:
\[ \frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2 + b^2} \]

### Final Locus Equation:
The locus of point \( P \) is:
\[ \frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2 + b^2} \]