Problem :
The point of intersection of the tangents to the parabola $y^2=4x$ at the points where the circle $(x-3)^2+y^2=9$ meets the parabola, other than the origin, is ..
Solution :
Point of the intersection of the parabola and circle is given by $(x-3)^2+(4x)^2=9$
$\Rightarrow x^2-6x +9 +16x^2=9$ $\Rightarrow 17x^2-6x=0 $ $\Rightarrow x(17x-6)=0$ $\Rightarrow x =0 ; x = 6/17$
$y =0, $ and $y =\sqrt{\frac{24}{17}}$
How to proceed to find the point of intersection of tangents please suggest. Thanks.