Find the equation of the parabola whose axis is horizontal, vertex on the y-axis and which passes through (2,4) and (8,-2).
the general formula for a parabola with known vertex and horizontal axis is
x = a(y-k)^2 + h, where (h,k) is the vertex
in your case the vertex , being on the y-axis , can be called (0,k) , and the equation would be
x = a(y - k)^2 + 0
for the point (2,4)
2 = a(4-k)^2 (#1)
for the point (8,-2)
8 = a(-2-k)^2 (#2)
divide #2 by #1
4 = (-2-k)^2 / (4-k)^2
4(4-k)^2 = (-2-k)^2
64 - 32k + 4k^2 = 4 + 4k + k^2
3k^2 -36 + 60 = 0
k^2 - 12 + 20 = 0
(k-10)(k-2) = 0
k = 10 or k = 2
if k=2, x = a(y-2)^2
and using (2,4)
2 = a(4-2)^2
a = 1/2 ---------> x = (1/2)(y-2)^2
if k=10 , x = a(y-10)^2
again using (2,4)
2 = a(4-10)^2
a = 1/18 ----------> x= (1/18)(y-10)^2
Notice that there are two possible equations,
btw, I checked both equations using the point(8,-2), it also satisfies both equations.
Why did the parabola see a therapist? Because it had trouble finding its vertex on the y-axis! But no worries, I'm here to help you out!
First, let's consider the general equation of a parabola with a horizontal axis, given by y = a(x - h)^2 + k. The vertex of this parabola would be (h, k), where h represents the x-coordinate and k represents the y-coordinate.
Since the parabola's vertex is on the y-axis, we know that h = 0 and k = 0. This simplifies the equation to y = ax^2.
Now, let's use the given points (2,4) and (8,-2) to find the value of a.
Substituting the coordinates of (2,4) into the equation, we get:
4 = a(2)^2
4 = 4a
a = 1
Substituting the coordinates of (8,-2) into the equation, we get:
-2 = a(8)^2
-2 = 64a
a = -1/32
Wait, something doesn't seem right! It appears that the values of a obtained from both points are different. This means that there is no single parabola that passes through both points and has its vertex on the y-axis.
I apologize if I couldn't provide the equation you were looking for. But hey, at least we had a little fun with the parabola's therapy session, right?
To find the equation of the parabola, we need to determine its standard form, which is given by the equation:
y = a(x - h)^2 + k
where (h, k) represents the coordinates of the vertex.
Since the axis of the parabola is horizontal and the vertex lies on the y-axis, the equation can be simplified as:
y = a(x - 0)^2 + k
y = ax^2 + k
Now we need to find the values of a and k.
Given that the parabola passes through (2,4), we can substitute these coordinates into the equation to get:
4 = a(2^2) + k
4 = 4a + k -- Eq. 1
Similarly, using the point (8,-2):
-2 = a(8^2) + k
-2 = 64a + k -- Eq. 2
Now we have a system of equations with two variables (a and k):
4 = 4a + k -- Eq. 1
-2 = 64a + k -- Eq. 2
To solve this system, we'll subtract Eq. 2 from Eq. 1:
4 - (-2) = (4a + k) - (64a + k)
Simplifying this equation:
6 = 4a - 64a
Combining like terms:
6 = -60a
Dividing both sides by -60:
a = -1/10
Now that we have the value of a, we can substitute it back into Eq. 1 or Eq. 2 to solve for k. Let's use Eq. 1:
4 = 4(-1/10) + k
4 = -4/10 + k
Combining like terms:
4 = -2/5 + k
To get rid of the fraction, we can rewrite 4 as 20/5:
20/5 = -2/5 + k
Now we have a common denominator:
20/5 = (-2 + 5k) / 5
Multiplying both sides by 5:
20 = -2 + 5k
Adding 2 to both sides:
20 + 2 = -2 + 2 + 5k
Simplifying:
22 = 5k
Dividing both sides by 5:
k = 22/5
Therefore, the equation of the parabola is:
y = (-1/10)x^2 + 22/5
To find the equation of a parabola, we need to use the general form of the equation for a parabola with a horizontal axis:
y = a(x - h)^2 + k
Here, (h, k) is the vertex of the parabola. Since the vertex of the parabola is on the y-axis, the x-coordinate of the vertex is 0. Therefore, we have h = 0.
Let's first find the value of k, the y-coordinate of the vertex. We know that the vertex lies on the y-axis, so it has the form (0, k). We need to find k.
Now we are given two points that lie on the parabola: (2, 4) and (8, -2). We can substitute these coordinates to find the values of k and a.
For the point (2,4), substituting x = 2 and y = 4 into the equation y = a(x - h)^2 + k, we get:
4 = a(2 - 0)^2 + k
4 = 4a + k ----(1)
Similarly, for the point (8,-2), substituting x = 8 and y = -2, we get:
-2 = a(8 - 0)^2 + k
-2 = 64a + k ----(2)
Now we have a system of two equations with two variables: a and k. We can solve this system to find the values of a and k.
Subtracting equation (1) from equation (2), we get:
-2 - 4 = 64a + k - (4a + k)
-6 = 60a
a = -1/10 ----(3)
Substituting this value of a back into equation (1), we can solve for k:
4 = 4(-1/10) + k
4 = -4/10 + k
4 = -2/5 + k
4 + 2/5 = k
k = 22/5 ----(4)
Therefore, the equation of the parabola is:
y = (-1/10)x^2 + 22/5