Name Cumulative Test Due Monday \( 3 / 17 / 25 \) 10. (4pt) Determine an equation for the parabola with focus. \( (-2,4) \) and directrix \( y=10 \). (The use of the grid below is optional.)

1. The definition of a parabola is the set of all points \((x,y)\) that are equidistant from the focus \((-2,4)\) and the directrix \(y=10\). Thus, for any point \((x,y)\) on the parabola, \[ \sqrt{(x+2)^2+(y-4)^2} = |y-10|. \] 2. Square both sides to eliminate the square root: \[ (x+2)^2+(y-4)^2 = (y-10)^2. \] 3. Expand each term: \[ \begin{aligned} (x+2)^2 &= x^2+4x+4, \\ (y-4)^2 &= y^2-8y+16, \\ (y-10)^2 &= y^2-20y+100. \end{aligned} \] Substituting these gives: \[ x^2+4x+4+y^2-8y+16 = y^2-20y+100. \] 4. Simplify the equation by canceling \( y^2 \) from both sides and combining like terms: \[ x^2+4x+4-8y+16 = -20y+100. \] Combine the constants \(4+16\) on the left side: \[ x^2+4x+20-8y = -20y+100. \] 5. Isolate \( y \) terms on one side: \[ x^2+4x+20-8y+20y-100 = 0, \] which simplifies to: \[ x^2+4x+12y-80 = 0. \] 6. An alternative approach is to find the vertex of the parabola. The vertex lies midway between the focus and the directrix. - The \(y\)-coordinate of the vertex is \[ k=\frac{4+10}{2}=7. \] - Since the focus has \(x=-2\), the vertex has \(h=-2\). The distance from the vertex to the focus is \[ p=7-4=3, \] and since the focus is below the vertex, we have \(p=-3\). In vertex form for a vertical parabola, \[ (x-h)^2=4p(y-k), \] substituting \(h=-2\), \(k=7\), and \(p=-3\) gives: \[ (x+2)^2=4(-3)(y-7)= -12(y-7). \] 7. The equation of the parabola is therefore: \[ (x+2)^2 = -12(y-7). \]