SOLUTION: How do you find the focus and the directrix of a parabola when the equation is (x-2)^2=2(y+2)?

Question 247673: How do you find the focus and the directrix of a parabola when the equation is (x-2)^2=2(y+2)?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
(x-2)^2=2(y+2)?

Your solution is at this link.

http://home.windstream.net/okrebs/page64.html

Your equation needs to be put in the right form.

The right form is:

(y-k) = +/- (1/4p) * (x - h)^2

Where:

(h,k) is the vertex of the parabola.
p is the distance from the focus to the vertex and from the vertex to the directrix.

your equation is:

(x-2)^2=2(y+2)

flip it around so you get:

2 * (y+2) = (x-2)^2

divide both sides by 2 so you get:

(y+2) = (1/2)*(x-2)^2

looks like your vertex is at (h,k) = (2,-2)

You have 1/(4*p) = 1/2

multiply both sides by 4*p and you get:

1 = (1/2) * 4p which becomes:

1 = 2*p which becomes

p = (1/2) after you divide both sides of the equation by 2.

Solve for y to be able to graph this equation.

(y+2) = (1/2)*(x-2)^2 becomes:

y = (1/2)*(x-2)^2 - 2 after you subtract 2 from both sides of the equation.

Graph of this equation looks like:

Your vertex is at the point (x,y) = (2,-2)

Your focus is at at the point (x,y) = (2,-3/2)

Your directrix is at the line (x,y) = (x,-5/2) which translates to the equation of the line as y = -5/2.

If this is the true focus and directrix, then the distance between the focus and any point on the graph of the equation of the parabola should be the same as the distance between any point on the graph of the equation of the parabola and a perpendicular line dropped from that point to a point on the line of the directrix.

you can see this is true at the point (2,-2) on the graph because the distance from (2,-2) to (2,-3/2) is the same as the distance from (2,-2) to (2,-5/2).

Another easy point to check is where a horizontal line from the focus intersects with the graph.

Based on your equation of ...

y = (1/2)*(x-2)^2 - 2, when y = -3/2, then x = 2 +/- sqrt(1).

This comes out to be x = 1 and x = 3.

Since the focus is at x = 2, this means that x is 1 unit away from the focus in either direction when y = -3/2.

Since the directrix is at y = -5/2, and the points on the graph are at (1,-3/2) and (3,-3/2), then each point is 1 unit away from the directrix because ((-5/2) - (-3/2)) = 1.

Your solution is:

Vertex is (x,y) = (h,k) = (2,-2)

p = 1/2 = distance from focus to vertex and distance from vertex to directrix when x = 2.

Focus is at (x,y) = (2,(-3/2))

Directrix is at (x,y) = (x,(-5/2))

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