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,-9Â
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 getÂ
$$2x+2y\frac{\text{d}y}{\text{d}x}=0$$
Slope of the tangent to the circle $$\ m_1= -1/ x$$
On differentiating the equation of the parabola ,we getÂ
$$2y\frac{\text{d}y}{\text{d}x}=8$$
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.
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