What is the equation of the parabola that has a vertex at $(1,2)$ and a directrix at $y$ $=-2$ ? $x^{2}=8(y+4)$ $(x-1)^{2}=16(y-2)$ $x^{2}=24(y-8)$ $y^{2}=12(x-4)$ $y^{2}=20(x-10)$
Solution
Step 1 ：The equation of a parabola can be written in the form \((x-h)^2=4p(y-k)\) if the parabola opens upwards or downwards, or in the form \((y-k)^2=4p(x-h)\) if the parabola opens to the left or right. Here, \((h,k)\) is the vertex of the parabola and \(p\) is the distance from the vertex to the focus or the directrix.
Step 2 ：Given that the vertex of the parabola is at \((1,2)\), we can substitute \(h=1\) and \(k=2\) into the equations.
Step 3 ：Also, since the directrix is at \(y=-2\), the distance \(p\) from the vertex to the directrix is \(2-(-2)=4\).
Step 4 ：Since the directrix is below the vertex, the parabola opens upwards. Therefore, the equation of the parabola should be in the form \((x-h)^2=4p(y-k)\).
Step 5 ：Substituting \(h=1\), \(k=2\), and \(p=4\) into this equation, we get \((x-1)^2=16(y-2)\).
Step 6 ：Therefore, the equation of the parabola that has a vertex at \((1,2)\) and a directrix at \(y=-2\) is \((x-1)^2=16(y-2)\).
Step 7 ：Final Answer: The equation of the parabola that has a vertex at \((1,2)\) and a directrix at \(y=-2\) is \(\boxed{(x-1)^2=16(y-2)}\).
