A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola tangentially at exactly 2 points. Find these 2 points.

Thanks!

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- November 17th 2007, 06:22 PMpolymeraseIntersection of Circle and Parabola
A circle with radius 1 has its centre on the positive y-axis. The circle touches the parabola tangentially at exactly 2 points. Find these 2 points.

Thanks! - November 17th 2007, 07:47 PMJhevon
Let the center of the circle be for

thus our two curves are:

and

and we want these two curves to have only two points in common. (it is okay to take this interpretation since the circle will be "inside" the parabola)

plug in into the equation of the circle, we get:

the above is quadratic in

thus, ..........................(1)

we want this to have only two solutions, therefore, set the discriminant to zero.

thus, it remains for us to solve:

for .

knowing the value for , we plug it into equation (1) to find the x-coordinate for the point of intersection. then use that to get the y-coordinate. (note, one of your solutions for y will be erroneous, be sure to make sure your solutions work in BOTH of the original equations). i leave the rest to you, it's pretty much following your nose from here - November 17th 2007, 08:19 PMkalagota
i have solved this for my tutor before but i can't find my solutions..

but i think, this can be solved in two ways, one is using calculus while the other is using some college algebra..

let (0,k), k>0 be the center of the circle:

let us use the fact that both are symmetric wrt to y axis.. hence, let (a,b) and (-a,b) be the points of intersection..

so, and and so if we combine both, we have

or

using quadratic formula, we have:

for that to exists, we need or

in fact, what we need is that so that we have a unique solution for and thathas 2 solutions.. Ü so if , we have or . and*a*

hence, the points are and

EDIT: i hang up to eat; and then, jhevon had answered it already.. hehe - November 17th 2007, 08:28 PMJhevon
- November 18th 2007, 06:38 AMpolymerase
- November 18th 2007, 06:39 AMpolymerase