To rewrite the given equation in the standard form of a parabola, we first identify the standard form for a parabola that opens horizontally:

$(y - k)^2 = 4p(x - h)$

where $(h, k)$ is the vertex of the parabola and $p$ determines the distance from the vertex to the focus.

Step-by-step Solution:

Given equation: $-y^2 + 8y + x + 12 = 0$

First, isolate $x$:

$x = y^2 - 8y - 12$

Next, we complete the square for the $y$ terms.

1. Complete the square:

   Take the coefficient of $y$, which is $-8$, halve it to get $-4$, and then square it to get $16$.

2. Rewrite the equation:

   Add and subtract 16 inside the equation:

   $x = (y^2 - 8y + 16) - 16 - 12$

   Simplify:

   $x = (y - 4)^2 - 28$

3. Rewrite in standard form:

   $(y - 4)^2 = x + 28$

Thus, the standard form of the given equation is:

$(y - 4)^2 = 4p(x + 28)$

where $4p = 1$. Therefore, $p = \frac{1}{4}$.

Final Answer:

$(y - 4)^2 = x + 28$

This represents a parabola that opens to the right with vertex at $(-28, 4)$.

Would you like more details or have any questions?

Related Questions:

1. What is the focus of this parabola?
2. How do we convert a horizontal parabola to a vertical one?
3. How is completing the square useful in deriving the standard form?
4. How can we determine the axis of symmetry from this equation?
5. How do you find the directrix of a parabola?

Tip:

When completing the square, always remember to balance the equation by adding or subtracting the same value on both sides!