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Math  /  Geometry

QuestionWhat is the focus of the parabola in the standard form equation (x+3)2=20(y−6)?(x+3)^{2}=20(y-6) ? (A) (6,2)(6,2) (B) (−3,11)(-3,11) (C) (2,6)(2,6) (D) (11,−3)(11,-3)

Studdy Solution

STEP 1

1. The given equation of the parabola is in the standard form (x−h)2=4p(y−k)(x-h)^2 = 4p(y-k).
2. The vertex of the parabola is at the point (h,k)(h, k).
3. The parameter pp determines the distance from the vertex to the focus.
4. The focus of the parabola can be found using the vertex and the value of pp.

STEP 2

1. Identify the vertex (h,k)(h, k) from the given equation.
2. Determine the value of pp from the given equation.
3. Use the vertex and pp to find the coordinates of the focus.

STEP 3

Rewrite the given equation (x+3)2=20(y−6)(x+3)^2 = 20(y-6) in the standard form (x−h)2=4p(y−k)(x-h)^2 = 4p(y-k).
By comparing, we see that: h=−3andk=6 h = -3 \quad \text{and} \quad k = 6

STEP 4

Identify the value of 4p4p from the given equation.
From (x+3)2=20(y−6)(x+3)^2 = 20(y-6), we have: 4p=20 4p = 20 Thus, p=204=5 p = \frac{20}{4} = 5

STEP 5

Determine the coordinates of the focus using the vertex (h,k)(h, k) and the value of pp.
Since the parabola opens upwards (as indicated by the (x−h)2(x-h)^2 form), the focus will be pp units above the vertex. Therefore, the coordinates of the focus are: (h,k+p)=(−3,6+5)=(−3,11) (h, k + p) = (-3, 6 + 5) = (-3, 11)
The focus of the parabola is (−3,11)(-3, 11), which corresponds to option (B).

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