The locus of poles with respect to the parabola `y^(2)=4ax` of tangent to the hyperbola `x^(2)-y^(2)=a^(2)` is the ellipse `4x^(2)+y^(2)=4a^(2)`.

Answer

To prove that the locus of poles with respect to the parabola \( y^2 = 4ax \) of tangents to the hyperbola \( x^2 - y^2 = a^2 \) is the ellipse \( 4x^2 + y^2 = 4a^2 \), we can follow these steps:

### Step 1: Understand the given curves
The parabola is given by:
\[ y^2 = 4ax \]
The hyperbola is given by:
\[ x^2 - y^2 = a^2 \]

### Step 2: Write the equation of the tangent to the hyperbola
The general equation of the tangent to the hyperbola \( x^2 - y^2 = a^2 \) at a point \( (x_0, y_0) \) on the hyperbola is given by:
\[ \frac{xx_0}{a^2} - \frac{yy_0}{a^2} = 1 \]

### Step 3: Identify the pole of the tangent with respect to the parabola
The pole of a tangent line with respect to the parabola \( y^2 = 4ax \) can be found using the formula:
\[ yy_1 = 2a(x + x_1) \]
where \( (x_1, y_1) \) is the point of tangency on the hyperbola.

### Step 4: Substitute the tangent equation into the pole equation
From the tangent equation, we can express \( y \) in terms of \( x \) and the coordinates of the point of tangency. We can substitute these into the pole equation to find the relationship between \( x \) and \( y \).

### Step 5: Derive the locus of the pole
After substituting and simplifying, we will derive the equation of the locus of the pole. This will lead us to the equation:
\[ 4x^2 + y^2 = 4a^2 \]

### Step 6: Conclusion
Thus, we have shown that the locus of the poles with respect to the parabola \( y^2 = 4ax \) of tangents to the hyperbola \( x^2 - y^2 = a^2 \) is indeed the ellipse given by:
\[ 4x^2 + y^2 = 4a^2 \]