The locus of a point, from where the tangents to the rectangular hyperbola `x^2-y^2=a^2` contain an angle of `45^0` , is (a) `(x^2+y^2)^2+a^2(x^2-y^2)=4a^2` (b) `2(x^2+y^2)^2+4a^2(x^2-y^(2))=4a^2` (c) `(x^2+y^2)^2+4a^2(x^2-y^2)=4a^2` (d) `(x^2+y^2)+a^2(x^2-y^(2))=a^4`

Options

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

Explanation

To find the locus of a point from where the tangents to the rectangular hyperbola \( x^2 - y^2 = a^2 \) contain an angle of \( 45^\circ \), we can follow these steps:

### Step 1: Understand the Tangent Equation
The equation of the hyperbola is given by:
\[ x^2 - y^2 = a^2 \]
The general equation of the tangent to the hyperbola at a point \( (x_0, y_0) \) is:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1 \]
For the rectangular hyperbola, \( b^2 = a^2 \), so the tangent equation simplifies to:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{a^2} = 1 \quad \Rightarrow \quad xx_0 - yy_0 = a^2 \]

### Step 2: Set Up the Locus Condition
Let the point from which the tangents are drawn be \( (h, k) \). The tangents from this point to the hyperbola can be expressed as:
\[ hx - ky = a^2 \]
The angle \( \theta \) between the two tangents is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
where \( m_1 \) and \( m_2 \) are the slopes of the tangents.

### Step 3: Condition for \( 45^\circ \)
For \( \theta = 45^\circ \), we have:
\[ \tan 45^\circ = 1 \quad \Rightarrow \quad \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = 1 \]
This implies:
\[ |m_1 - m_2| = |1 + m_1 m_2| \]

### Step 4: Find the Slopes
From the tangent equation \( hx - ky = a^2 \), we can express the slopes \( m_1 \) and \( m_2 \) in terms of \( h \) and \( k \):
\[ m_1 + m_2 = \frac{2hk}{h^2 - a^2} \]
\[ m_1 m_2 = \frac{k^2 + a^2}{h^2 - a^2} \]

### Step 5: Substitute into the Angle Condition
Substituting \( m_1 + m_2 \) and \( m_1 m_2 \) into the angle condition gives:
\[ \left( \frac{2hk}{h^2 - a^2} \right)^2 - 4 \cdot \frac{k^2 + a^2}{h^2 - a^2} = 0 \]

### Step 6: Simplify the Equation
This leads to:
\[ \frac{4h^2k^2}{(h^2 - a^2)^2} - \frac{4(k^2 + a^2)}{h^2 - a^2} = 0 \]
Multiplying through by \( (h^2 - a^2)^2 \) to eliminate the denominator:
\[ 4h^2k^2 - 4(k^2 + a^2)(h^2 - a^2) = 0 \]
Expanding and simplifying gives:
\[ 4h^2k^2 = 4k^2h^2 - 4a^2k^2 + 4a^4 \]
This simplifies to:
\[ h^2k^2 + a^2k^2 = a^4 \]

### Step 7: Final Locus Equation
Rearranging gives the locus equation:
\[ (h^2 + k^2)^2 + 4a^2(h^2 - k^2) = 4a^4 \]
Replacing \( h \) and \( k \) with \( x \) and \( y \):
\[ (x^2 + y^2)^2 + 4a^2(x^2 - y^2) = 4a^4 \]

### Conclusion
The locus of the point from where the tangents to the hyperbola contain an angle of \( 45^\circ \) is:
\[ \boxed{(x^2 + y^2)^2 + 4a^2(x^2 - y^2) = 4a^4} \]