The circle $$x^2 + y^2 = 8$$ intersects the parabola $$y^2 = 2x$$ 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

First, we will try and find the point of intersection of the two curves.

We do so by equating the two curves.

So, $$x^2+\ y^2-8=y^2-2x$$

=> $$x^2+2x\ -8=0$$

=> (x+4)(x-2)=0

=> x= -4 or 2.

But since, the point of intersection lies on the first quadrant, x=2.

By putting the value of x in any of the two curves, we can find the value of y. 

'.' $$y^2=2x$$

=> $$y^2=4$$

=> y= 2 or -2. But the point of intersection is in first quadrant, so y=2.

So, P= (2,2)

Now, for the parabola $$y^2=2x$$, tangent can be found by differentiating the curve at (2,2).

=> $$2y\ \frac{\ dy}{dx}=2$$

=>$$\frac{\ dy}{dx}=\ \frac{\ 1}{y}$$

=> $$\frac{\ dy}{dx}=\ \frac{\ 1}{2}$$

.'. Angle made by the tangent to parabola at P, tan$$\alpha\ $$= $$\ \frac{\ 1}{2}$$.

Similarly, tangent at the circle $$x^2+y^2=8$$ can be found by differentiating the curve at (2,2)

=> $$2x+2y\ \frac{\ dy}{dx}=0$$

=> $$2y\ \frac{\ dy}{dx}=-2x$$ 

=> $$\frac{\ dy}{dx}=\ \frac{\ -x}{y}$$

=> $$\frac{\ dy}{dx}=\ -1$$.

.'. Angle made by the tangent to circle at P, tan$$\beta\ $$= -1

Now, the acute angle between the two tangents is represented by $$\alpha\ -\beta\ $$.

We will use the expression $$\tan\left(\alpha-\beta\right)\ =\ \ \frac{\ \tan\ \alpha\ -\tan\ \beta\ }{1+\tan\ \alpha\ \tan\ \beta\ }$$

=> $$\tan\left(\alpha\ -\beta\ \right)=\ \ \frac{\ \ \frac{\ 1}{2}-\left(-1\right)}{1-\frac{\ 1}{2}}$$

=>$$\tan\left(\alpha\ -\beta\ \right)=\ \ \frac{\ \ \ \frac{\ 3}{2}}{\ \ \frac{\ 1}{2}}$$

=>$$\tan\left(\alpha\ -\beta\ \right)=\ \ 3$$

.'. $$\alpha\ -\beta\ =\ \ \tan^{-1}\left(3\right)$$. therefore, option A