the parabola with equation y=(x-4)^2 +3 has its vertex at (4,3) and passes through (5,4). Find an equation of a different parabola with its vertex at (4,3) that passes through (5,4)
2007-04-22 05:14:01 · 6 answers · asked by Mark 2 in Science & Mathematics ➔ Mathematics
GIVEN:
(y-3) = (x-4)² that passes through (5,4)
HOW ABOUT:
(y-3) = - (x-4)² ?
.....no I don't think so.
That wouldn't do at all. The parabola would never pass through (5,4).
It can't be a vertical parabola...
...to have the same vertex it must go the opposite direction, but then it wouldn't go through the point (5,4).
THEREFORE it must be a Horizontal parabola with "y" squared, not "x".
....and the equation must open to the right to pass through the point (5,4) on the other parabola.
FINAL ANSWER: (x-4) = (y-3)²
2007-04-22 05:21:07 · answer #1 · answered by LovesMath 3 · 0⤊ 0⤋
The parabola which satisfies this condition would have to be one which is perpendicular to the first, so it would have to be in the form x = (y - a')² + b', where a' = 3 and b' = 4. Note that since y is positive, x must also be positive. Otherwise, the second parabola would open to the left and would not pass through the proper branch of the first parabola to coincide with the point (5,4), which lies on the right branch. Hence, the equation of the new parabola is:
x = (y - 3)² + 4.
Now, let's test this. If we let x = 5, then (y - 3)² = 5 - 4 = 1.
Then y² - 6y + 9 = 1 ----> y² - 6y + 8 = 0.
The above equation factors into:
(y - 2)(y - 4) = 0.
So (y - 2) = 0 or (y - 4) = 0. This implies y = 2 or 4.
We reject y = 2, because it lies on the bottom branch of the second parabola, and does not satisfy our conditions.
So y = 4, and the equation of our new parabola is indeed x = (y - 3)² + 4.
.
2007-04-22 13:07:14 · answer #2 · answered by MathBioMajor 7 · 1⤊ 0⤋
One possible equation is x = (y - 3)^2 + 4
There are others where the axis of the parabola is at an angle to the coordinate axes.
Edit. The answer two below is the same parabola as the one given at the start. (x - 4)^2 = (4 - x)^2
2007-04-22 12:19:31 · answer #3 · answered by mathsmanretired 7 · 2⤊ 0⤋
If the vertex is at (4,3) the parabola looks like
y = a(x-4)² + 3
Now plug in your point
4 = a(1) + 3,
so a must be 1.
There is no other parabola.
You can see from this that a parabola is uniquely
determined by its vertex and one other point on it.
2007-04-22 12:30:02 · answer #4 · answered by steiner1745 7 · 0⤊ 2⤋
y = (-x + 4)^2 + 3
2007-04-22 12:24:59 · answer #5 · answered by sweetwater 7 · 0⤊ 2⤋
tried for about ten minutes to solve. have no answer and don't think there is one but can't prove it.
2007-04-22 12:32:00 · answer #6 · answered by verner66 2 · 0⤊ 2⤋