QuestionJuly 15, 2025

Write an equation for the following conic section Parabola with vertex (-1,2) and focus (2,2)

Write an equation for the following conic section Parabola with vertex (-1,2) and focus (2,2)
Write an equation for the following conic section Parabola with vertex (-1,2) and focus (2,2)

Solution
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Answer

(y-2)^2 = 12(x+1) Explanation 1. Determine the orientation The focus (2,2) and vertex (-1,2) have the same y-coordinate, indicating a horizontal parabola. 2. Calculate the distance between vertex and focus The distance p is |2 - (-1)| = 3. Since the focus is to the right of the vertex, p = 3. 3. Write the equation of the parabola For a horizontal parabola with vertex (h,k) and focus (h+p,k), the equation is (y-k)^2 = 4p(x-h). Here, (h,k) = (-1,2) and p = 3, so **(y-2)^2 = 12(x+1)**.

Explanation

1. Determine the orientation<br /> The focus $(2,2)$ and vertex $(-1,2)$ have the same $y$-coordinate, indicating a horizontal parabola.<br /><br />2. Calculate the distance between vertex and focus<br /> The distance $p$ is $|2 - (-1)| = 3$. Since the focus is to the right of the vertex, $p = 3$.<br /><br />3. Write the equation of the parabola<br /> For a horizontal parabola with vertex $(h,k)$ and focus $(h+p,k)$, the equation is $(y-k)^2 = 4p(x-h)$. Here, $(h,k) = (-1,2)$ and $p = 3$, so **$(y-2)^2 = 12(x+1)$**.
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