The locus of a point from which two tangent are drawn to `x^2-y^2=a^2` which are inclined at angle `(pi)/(4)` to each other is

Options

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

Answer

To find the locus of a point from which two tangents are drawn to the hyperbola \(x^2 - y^2 = a^2\) that are inclined at an angle of \(\frac{\pi}{4}\) to each other, 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 tangents drawn from a point \((h, k)\) to this hyperbola is:
\[ x h - y k - a^2 = 0 \]

### Step 2: Use the Condition for Tangents
The condition for two tangents from a point to be inclined at an angle \(\theta\) is given by:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
For our case, since the angle is \(\frac{\pi}{4}\), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, we can set up the equation:
\[ 1 = \frac{2\sqrt{h^2 - a^2}}{a + a} = \frac{2\sqrt{h^2 - a^2}}{2a} \]
This simplifies to:
\[ 1 = \frac{\sqrt{h^2 - a^2}}{a} \]

### Step 3: Square Both Sides
Squaring both sides gives:
\[ 1 = \frac{h^2 - a^2}{a^2} \]
Multiplying through by \(a^2\) results in:
\[ a^2 = h^2 - a^2 \]
Rearranging gives:
\[ h^2 = 2a^2 \]

### Step 4: Substitute for Locus
Since \(h\) and \(k\) represent the coordinates of the point from which tangents are drawn, we can replace \(h\) with \(x\) and \(k\) with \(y\):
\[ x^2 = 2a^2 \]

### Step 5: Final Locus Equation
Thus, we have:
\[ x^2 - 2a^2 = 0 \]
This represents a vertical line in the coordinate plane.

### Conclusion
The locus of the point from which two tangents are drawn to the hyperbola \(x^2 - y^2 = a^2\) that are inclined at an angle of \(\frac{\pi}{4}\) to each other is given by:
\[ x^2 = 2a^2 \]