# Another Parabola help

• Jan 20th 2007, 06:40 AM
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 :)
• Jan 20th 2007, 09:02 AM
CaptainBlack
Quote:

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
• Jan 20th 2007, 09:37 AM
ticbol
Quote:

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.
• Jan 20th 2007, 10:11 AM
Soroban

Quote:

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.

• Jan 21st 2007, 06:18 AM