The straight line y=x+3 cuts the circle x^2+y^2-6x-4y-3=0 at points P and Q, where Q lies in the first quadrant.
Calculate the coordinates of P and Q. Then, state the coordinates of the point of intersection of the tangents to the circle at points P and Q.
The midpoint of PQ is the vertex of a parabola with its axis parallel to the y-axis. Given that the parabola also passes through the point P, find the equation of the parabola and state the coordinates of its focus.

Question

The straight line y=x+3 cuts the circle x^2+y^2-6x-4y-3=0 at points P and Q, where Q lies in the first quadrant.
	Calculate the coordinates of P and Q. Then, state the coordinates of the point of intersection of the tangents to the circle at points P and Q.
	The midpoint of PQ is the vertex of a parabola with its axis parallel to the y-axis. Given that the parabola also passes through the point P, find the equation of the parabola and state the coordinates of its focus.

anonymous · Answer

$$x^2+y^2-6x-4y-3=0$$
$$x^2+(x+3)^2-6x-4(x+3)-3=0$$ solve for x

anonymous · Answer

$$2 x^2-4 x-6 = 0$$
$$2 (x-3) (x+1) = 0$$
$$x=3,x=-1$$

anonymous · Answer

when x=3,y=6 when x=-1,y=2
right?

anonymous · Answer

by considering point Q at the first quadrant you will know which point is Q and which point is P

anonymous · Answer

yes you are in first quadrant, so it is (3,6)

anonymous · Answer

oh it says "points P and Q" so Q is (3,6) P is (-1,2)

anonymous · Answer

the centre will (3,2).right?

anonymous · Answer

midpoint of the two is the average of the coordinates M = (1,4)

anonymous · Answer

the parabola will be open downward so he general equation will be -4a(y-k)=(x-h)2 and point (1,4) will be the (h,k)