Question

the parabola for A(2,3)B(-1,0)C(1,0)

Answer 1

The vertex of the parabola is the midpoint of the segment AB, which is (-1/3, 3/2).
Since the parabola has a vertical axis of symmetry, its equation is of the form:
(x - h)^2 = 4p(y - k)
where (h,k) is the vertex and p is the distance from the vertex to the focus.
We can use point C to find p:
(1 -(-1/3))^2 = 4p(0 - 3/2)
16/9 = -6p
p = -8/3
So the equation of the parabola is:
(x + 1/3)^2 = -32/3(y - 3/2)

Answer 2

WHAT ?
I think the midpoint of BC (not of AB) is on the axis of symmetry which is at the origin (0,0)

and by the way the irrelevant midpoint of AB is at (1/2 , 3/2) not (-1/3 , 3/2)

Answer 3

You're correct, I apologize for my mistake. The midpoint of BC is indeed on the axis of symmetry, which is the y-axis in this case. The midpoint of BC is (0, 0), so the vertex of the parabola is at the origin.

To find the focus of the parabola, we can use the distance formula. The distance from the vertex to point A is the same as the distance from the vertex to the directrix, which is a horizontal line p units below the vertex. The value of p can be found using the distance formula and the fact that the directrix is equidistant from points B and C:

p = (1/2) * sqrt((2 - (-1))^2 + (3 - 0)^2) = sqrt(13)/2

So the equation of the parabola is:

y^2 = 2px

Substituting p = sqrt(13)/2, we get:

y^2 = sqrt(13)x

Or, equivalently:

x = (1/sqrt(13))y^2

Answer 4

The vertex is NOT at the origin, the axis of symmetry is the y axis but the vertex is below y = 0

Answer 5

You're absolutely right, I apologize for my mistake. The vertex is indeed on the y-axis but below y=0.

To find the coordinates of the vertex, we can take the average of the y-coordinates of points B and C, which gives:

(0, (0+0)/2) = (0,0)

So, the vertex is at (0,0).

Using the distance formula and the fact that the distance from the vertex to point A is equal to the distance from the vertex to the directrix, we can find the equation of the parabola in standard form:

The distance from the vertex to point A is:

sqrt((2-0)^2 + (3-0)^2) = sqrt(13)

The distance from the vertex to the directrix is also sqrt(13). In this case, the directrix is the line y=-sqrt(13), which is p units below the vertex. Therefore, the equation of the parabola is:

y^2 = 4px

Substituting p=sqrt(13), we get:

y^2 = 4sqrt(13)x

Or, in vertex form:

x = (1/4sqrt(13))y^2

Answer 6

Agggrrhhh
(x - h)^2 = 4 a(y-k)
(h, k) is vertex (here we know h = 0)
x^2 = 4 a (y-k)
then use our points A(2,3)B(-1,0)C(1,0)
2^2 = 4 a ( 3-k) ---> 1 = 3 a - k a or a = 1/(3-k)
1 = 4 a (0-k) ---> a = -1/ (4k)
so
-1 /(4 k) = 1 /(3-k)
4 k = -3 + k
3 k = -3
k = -1
so
vertex at (0,-1

Answer 7

I apologize for the confusion, and thank you for correcting me. You are absolutely right. Using the standard form of the equation of a parabola, we can use the given points to find the value of a, and then use the vertex form to determine the coordinates of the vertex.

Starting with the equation:

(x - h)^2 = 4a(y - k)

We can substitute the coordinates of point A to get:

(2 - 0)^2 = 4a(3 - k)

Simplifying:

4 = 12a - 4ak

Dividing both sides by 4a gives:

1 = 3a - ak

We can also substitute the coordinates of point B:

(-1 - 0)^2 = 4a(0 - k)

1 = -4ak

Next, we can solve the system of equations:

1 = 3a - ak

1 = -4ak

From the second equation, we get:

a = -1/(4k)

Substituting this into the first equation, we get:

1 = 3(-1/(4k)) - (-1/(4k))k

Simplifying:

1 = (-3/4) + (1/4)k

Multiplying both sides by 4k gives:

4k = -3 + k

Solving for k:

3k = -3

k = -1

Substituting k = -1 into the equation a = -1/(4k), we get:

a = -1/(-4)

a = 1/4

Finally, substituting h = 0, k = -1, and a = 1/4 into the vertex form:

x = (1/(4p))(y - (-1))^2

We get:

x = (1/4)(y + 1)^2

Thus, the vertex of the parabola is at the point (0,-1).

Answer 8

so
x^2 = 4 a (y + 1)
try ( 2, 3)
4 = 4 a (4)
a = 1/4
so in the end
x^2 = 4(1/4) (y+1)
x^2 = y + 1

that works !

Answer 9

Yes, that's correct! The equation of the parabola is:

x^2 = y + 1

And we have verified that point A(2,3) satisfies this equation.

I apologize again for the confusion earlier. Thank you for correcting me and guiding me through the correct solution.

Answer 10

Whew , heavens to Betsy. I hope the student is still awake !

Answer 11

Yes, it's important to be clear and accurate in our explanations, to make sure the student understands the concepts and solutions properly. Thank you again for your help in correcting my mistakes.