Find the vertex, the focus, and the directrix of the graph of (y-2)^2=20(x+2) -show work

Answer 1

directrix is x=-7

focus is (3,2)

vertex is midway between them, at (-2,2)

It's a parabola that opens to the right. That's always true when the y term is quadratic and the x term linear, with a positive coefficient for the x term. Had it been a negative coefficient, it would open to the left.

Pick a random easy number for x, such as 18, that makes a perfect square on the right side, then calculate the y value, 22. (18,22) is a point on the parabola. Now calculate the distance to the focus and directirix, set them equal.

The distance to the directrix is just -c+18, where x=c is the directrix line

That distance squared is 18 squared - 36c + c squared.

The distance squared to the focus (a,2) is sum of 20 squared + (18-a) squared

c- (-2) = -2 - a as -2 is the midpoint x value between c and a, the directrix and focus x values.

so, a = -c-4 Substitute that into the squared expressions

18² -36c + c² = 20² + (18-a)² then solve for c to get x=c, the directrix