Let's go through each problem step-by-step.

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Problem 1

Find the equation of the parabola with vertex on the y-axis, axis of the parabola parallel to the x-axis, and passing through points $(2, 2)$ and $(8, -1)$.

Since the vertex is on the y-axis and the axis is parallel to the x-axis, the general form of the parabola is: $y = a(x - h)^2 + k$ where $(h, k)$ is the vertex. Given that the vertex is on the y-axis, $h = 0$, so the equation simplifies to: $y = ax^2 + k$

Since the parabola passes through $(2, 2)$ and $(8, -1)$, we can substitute these points into the equation to find $a$ and $k$.

1. For $(2, 2)$: $2 = a(2)^2 + k \Rightarrow 2 = 4a + k$

2. For $(8, -1)$: $-1 = a(8)^2 + k \Rightarrow -1 = 64a + k$

Now, we have the system of equations: $4a + k = 2$ $64a + k = -1$

Subtract the first equation from the second to eliminate $k$: $(64a + k) - (4a + k) = -1 - 2$ $60a = -3$ $a = -\frac{1}{20}$

Substitute $a = -\frac{1}{20}$ back into the first equation to find $k$: $4\left(-\frac{1}{20}\right) + k = 2$ $-\frac{1}{5} + k = 2$ $k = 2 + \frac{1}{5} = \frac{11}{5}$

Thus, the equation of the parabola is: $y = -\frac{1}{20}x^2 + \frac{11}{5}$

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Problem 2

Find the equation of the parabola with vertex on the line $y = 2$, axis parallel to the $y$-axis, length of the latus rectum is 6, and passing through $(2, 8)$.

Since the axis is parallel to the $y$-axis, the parabola has the form: $(x - h)^2 = 4p(y - k)$ where $(h, k)$ is the vertex, and the length of the latus rectum is $4|p|$.

1. Since the length of the latus rectum is 6, we have $4|p| = 6 \Rightarrow |p| = \frac{3}{2}$.

2. The vertex lies on the line $y = 2$, so $k = 2$.

3. To determine $h$ and the correct value of $p$, we use the fact that the parabola passes through $(2, 8)$.

Let’s assume the vertex is $(h, 2)$. Then the equation becomes: $(x - h)^2 = 4p(y - 2)$

Since the point $(2, 8)$ lies on the parabola, we substitute $x = 2$ and $y = 8$: $(2 - h)^2 = 4p(8 - 2)$ $(2 - h)^2 = 4p \cdot 6 = 24p$

Since $p$ can be $\pm \frac{3}{2}$, we have two possible cases to check.

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Problem 3

Find the equation of a parabola with axis parallel to the $x$-axis and passing through points $(5, 4)$, $(11, 2)$, $(21, -4)$.

Since the axis is parallel to the $x$-axis, the equation of the parabola is of the form: $y = ax^2 + bx + c$

We substitute each point into this equation to create a system of equations.

1. For $(5, 4)$: $4 = a(5)^2 + b(5) + c \Rightarrow 4 = 25a + 5b + c$

2. For $(11, 2)$: $2 = a(11)^2 + b(11) + c \Rightarrow 2 = 121a + 11b + c$

3. For $(21, -4)$: $-4 = a(21)^2 + b(21) + c \Rightarrow -4 = 441a + 21b + c$

Now we solve this system of equations for $a$, $b$, and $c$.

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Problem 4

Find the equation of a parabola with vertex on the line $y = x + 2$, axis parallel to the $y$-axis, latus rectum 6, and passing through $(-3, -1)$.

Since the axis is parallel to the $y$-axis, the equation has the form: $(x - h)^2 = 4p(y - k)$

1. Given the latus rectum length is 6, $4|p| = 6$, so $|p| = \frac{3}{2}$.
2. The vertex lies on the line $y = x + 2$, so $k = h + 2$.

Now, substitute $(-3, -1)$ into the equation to determine the specific values for $h$ and $p$.