# Thread: Another Parabola help

1. ## Another Parabola help

Hello, I Got a problem with this question im tryin to solve
and this is involved wth fractions which confuses me alot
Problem:

Axis VERTIcal;
Passes Through (-1,-3),(1,-2), & (2,1)
EDIT: Very Sorry For the problem!

The Answer at back of book is:
$\displaystyle 5x^2 + 3x - 6y - 20 = 0$

Thanks and very appreciated

2. Originally Posted by ^_^Engineer_Adam^_^
Hello, I Got a problem with this question im tryin to solve
and this is involved wth fractions which confuses me alot
Problem:

Axis Horizontal;
Passes Through (-1,-3),(1,-2), & (2,1)
EDIT: Very Sorry For the problem!

The Answer at back of book is:
$\displaystyle 5x^2 + 3x - 6y - 20 = 0$

Thanks and very appreciated
First observation: the answer in the back of the book is a parabola with
its axis vertical.

RonL

3. Originally Posted by ^_^Engineer_Adam^_^
Hello, I Got a problem with this question im tryin to solve
and this is involved wth fractions which confuses me alot
Problem:

Axis Horizontal;
Passes Through (-1,-3),(1,-2), & (2,1)
EDIT: Very Sorry For the problem!

The Answer at back of book is:
$\displaystyle 5x^2 + 3x - 6y - 20 = 0$

Thanks and very appreciated
Axis, horizontal.
Then the horizontal parabola is in the form
x = A*y^2 +B*y +C -------------------------(i)
But solving it that way, we get
5y^2 +y +12x -30 = 0 --------------answer
Which is not what the back of the book says.

--------------------------------------------------------------
So I tried Axis, vertical.
The form then is
y = A*x^2 +B*x +C -------------------------(ii)

At point (-1,-3):
-3 = A -B +C ---------------(1)

At point (1,-2):
-2 = A +B +C ---------------(2)

At point (2,1):
1 = 4A +2B +C --------------(3)

Solve those 3 equations simultaneously you'd get
A = 5/6
B = 1/2
C = -10/3
So, plugging those into (ii),
y = (5/6)x^2 +(1/2)x -10/3
Clear the fractions, multiply both sides by 6,
6y = 5x^2 +3x -20
Therefore,
0 = 5x^2 +3x -6y -20
5x^2 +3x -6y -20 = 0 --------------answer.

And the back of the book is happy.

Axis Horizontal
Passes Through (-1,-3), (1,-2), and (2,1)

The Answer at back of book is: $\displaystyle 5x^2 + 3x - 6y - 20 = 0$ . Impossible!

Their equation has $\displaystyle x^2$ and $\displaystyle y$ ... a parabola with a vertical axis.

5. Sory for the problem!
its given is vertical
And Thanks a lot for answering!