Question 1154423
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The distance from (2,6) to the focus (-1,2) is, by the Pythagorean Theorem, 5; and the (vertical) distance from (2,6) to the directrix y=1 is also 5.  So the point (2,6) is on the parabola.<br>
We can easily find 5 points on the parabola to enable us to make a good sketch of the graph.<br>
The focus is above the directrix, so the parabola opens upward.  With the parabola opening upward with the focus at (-1,2), the axis of symmetry of the parabola is x=-1.<br>
The vertex is the point on the axis of symmetry halfway between the directrix and focus -- at (-1,1.5).  That is our first point.<br>
We have shown that the point (2,6) is on the parabola; by symmetry the point (-4,6) is also on the parabola.  Those are our second and third points.<br>
For the last two points, we move right or left from the focus a distance equal to the distance between the focus and directrix.  Those two points will clearly be equally distant from the focus and directrix.  So our last two points are (2,2) and (0,2).<br>
Note those last two points are the endpoints of the latus rectum of the parabola -- the line segment perpendicular to the axis of symmetry (parallel to the directrix) passing through the focus.<br>
The vertex form of the equation of an upward opening parabola is<br>
{{{(y-k) = (1/(4p))(x-h)^2}}}<br>
where the vertex is (h,k) and p is the distance from the directrix to the vertex and from the vertex to the focus.<br>
The distance between the directrix and focus is 2p; in this example that distance is 1, so p is 1/2.<br>
Note that the 4p in the formula is the length of the latus rectum.<br>
We now know the vertex (h,k) is (-1,1.5); and we know p is 1/2 and so 4p is 2.  So we are ready to plug those numbers into the general formula to get the equation of this parabola.<br>
{{{(y-1.5) = (1/2)(x+1)^2}}}<br>
A graph....<br>
{{{graph(400,400,-6,6,-2,10,.5(x+1)^2+1.5)}}}<br>