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

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 \[ \boxed{(x+2)^2 = -12(y-7)} \]