The circle $$x^2 + y^2 = 9$$ intersects with the parabola $$y^2 = 8x$$ at a point P in the first quadrant. The acute angle between the tangents to the circle and the parabola at the point P is

Let's find the point of intersection of the circle $$x^2 + y^2 = 9$$ and the parabola $$y^2 = 8x$$

$$x^2 + 8x= 9$$

On solving the equation we will get the values of x as 1,-9 

Since the angle between the tangents has to be found in the first Quadrant, value of x is 1

$$\therefore value of y = 2\sqrt{2}$$

Let's find the slope of the tangent to the circle and the parabola

On differentiating the equation of the circle ,we get 

$$2x+2y\frac{\text{d}y}{\text{d}x}=0$$

Slope of the tangent to the circle $$\ m_1= -1/ x$$

On differentiating the equation of the parabola ,we get 

$$2y\frac{\text{d}y}{\text{d}x}=8$$

Slope of the tangent to the parabola $$m_2= 4/ y$$

Angle between the tangents $$\tan \theta$$= $$\frac{m_1-m_2}{1+m_1m_2}$$

                                           =$$\frac{-1/2\sqrt{2}-4/2\sqrt{2}}{1+-1/2\sqrt{2}*4/2\sqrt{2}}$$

                                           =$$\tan^{-1} (\frac{5}{\sqrt{2}})$$

Hence C is the correct answer.