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(4pt) Petermine an equation for the parabola with focus \( (-2,4) \) and directrix \( y=10 \). The of the grid below is optional.)

Ask by Bernard Luna. in the United States
Mar 16,2025

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The equation of the parabola is \( (x+2)^2 = -12(y-7) \).

Solution

We start with the definition of a parabola: it is the set of all points \( (x,y) \) that are equidistant from the focus and the directrix. Given the focus \( (-2,4) \) and directrix \( y=10 \), a point \( (x,y) \) on the parabola satisfies \[ \sqrt{(x+2)^2+(y-4)^2} = |y-10|. \] ### Step 1. Square both sides Squaring both sides of the equation to eliminate the square root gives \[ (x+2)^2 + (y-4)^2 = (y-10)^2. \] ### Step 2. Expand and Simplify Expand the terms: \[ \begin{aligned} (x+2)^2 &= x^2 + 4x + 4, \\ (y-4)^2 &= y^2 - 8y + 16, \\ (y-10)^2 &= y^2 - 20y + 100. \end{aligned} \] Substitute these into the equation: \[ x^2 + 4x + 4 + y^2 - 8y + 16 = y^2 - 20y + 100. \] ### Step 3. Cancel and Rearrange Terms Cancel \( y^2 \) from both sides: \[ x^2 + 4x - 8y + 20 = -20y + 100. \] Bring all terms to one side: \[ x^2 + 4x - 8y + 20 + 20y - 100 = 0, \] which simplifies to \[ x^2 + 4x + 12y - 80 = 0. \] ### Step 4. Write in Standard Form To express the parabola in standard form, complete the square for the \( x \)-terms: 1. Group the \( x \)-terms: \[ x^2 + 4x = (x^2 + 4x + 4) - 4 = (x+2)^2 - 4. \] 2. Substitute back into the equation: \[ (x+2)^2 - 4 + 12y - 80 = 0. \] 3. Combine the constant terms: \[ (x+2)^2 + 12y - 84 = 0. \] 4. Solve for \( y \): \[ 12y = 84 - (x+2)^2, \] \[ y = 7 - \frac{1}{12}(x+2)^2. \] Alternatively, the standard (vertex) form of a vertical parabola is \[ (x - h)^2 = 4p(y-k), \] where \( (h,k) \) is the vertex and \( p \) is the distance from the vertex to the focus. The vertex lies midway between the focus \( (-2,4) \) and the directrix \( y=10 \). The \( y \)-coordinate of the vertex is \[ k = \frac{4+10}{2} = 7, \] and its \( x \)-coordinate is the same as that of the focus \( h=-2 \). The distance \( p \) is \[ p = 4 - 7 = -3 \quad (\text{since the parabola opens downward}). \] Thus, the equation in vertex form is \[ (x+2)^2 = -12(y-7). \] ### Final Answer \[ \boxed{(x+2)^2 = -12(y-7)} \]

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To find the equation of a parabola given its focus and directrix, we start by identifying the vertex. The vertex is located halfway between the focus and the directrix. Hence, the y-coordinate of the vertex can be found by averaging the y-coordinates of the focus (4) and the directrix (10). Therefore, the vertex is at \( (-2, \frac{4 + 10}{2}) = (-2, 7) \). The distance \( p \) from the vertex to the focus or directrix is \( 10 - 7 = 3 \) (since the parabola opens downwards, \( p \) is negative). Parabolas with vertical axes of symmetry can be given by the equation \( (x - h)^2 = 4p(y - k) \), where \( (h,k) \) is the vertex. Plugging in values, we have: \[ (x + 2)^2 = -12(y - 7) \] This is the equation of the parabola with a focus at \( (-2, 4) \) and directrix \( y=10 \). And voilà! You’ve got the equation! Just remember, if you ever need to sketch this parabola, plotting the focus and the directrix can help you see how the curve behaves. The distance from any point on the parabola to the focus equals the distance from that point to the directrix, making it a perfect geometric shape. Happy graphing!

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