Math Is Fun Forum

Ultima Black Gate

Member

Registered: 2006-09-18

Posts: 14

Solution of this conic section problem

Hi guys I got this problem at a test and I couldnt solve it....

Find the equation of a parabola of vertical axis which has the points A(1,8) B(-2,-1) C(-3,8). Find the coordinates of the vertex.

Well heres what I did, I know that the formula of the parabola of center (h,k) is [(x-h)^2=4p(y-k)] so i noticed that points A & C have the same y-coordenate, so the right part of the eq would be equal. I then wrote (1-h)^2 = ( -3-h)^2 and from there I got a value of h.
Then I replaced that value on the other system of eqs I made with A & B to get k. After that I had p as the last unknown and it was easy to find on a third sys of 2 eqs.

Maybe I made some wrong mental calculations but my answer was not ok, and I wanted to know If at least the algorithm I tried to use is ok.

Thanx in advance!

Last edited by Ultima Black Gate (2007-12-12 02:44:12)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Solution of this conic section problem

I don't know your method, but here's another way.
vertex is at (-1, ?), where ? is less than -1.

Horizontal distances from vertex: 1²=1 and  2²=4
Subtract them, and get 4 - 1 = 3, Multiply 3 by 3 to get 9.
9 is 8 - (-1).

So the parabola is 3 times higher than y = x² .

So moving from (-2,-1) to the right one unit and down 3 units,
we get the vertex at (-1, -4).

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igloo myrtilles fourmis

Ultima Black Gate

Member

Registered: 2006-09-18

Posts: 14

Re: Solution of this conic section problem

The answer is correct.Thanks Mr. Franklin.