what is the equation of the parabola with a focus of (-2,5) and a directrix at y=3

Answer 1

y=(1/4)(x+2)^2 +4

Answer 2

On scratch graph paper, plot the directrix, y = 3. Plot the focus (-2,5). Since the directrix is horizontal and the focus is above the directrix, this is an upward opening parabola. Sketch the axis of symmetry as the line x = -2. The vertex lies on the axis of symmetry and is the midpoint of the segment with endpoints (-2,5) and (-2,3). Not that (-2,3) is simply the point of intersection of the directrix and the axis of symmetry. The vertex is therefore located at (-2,4). The letter p is usually used to represent the directed distance from the vertex to the focus. Since the parabola opens upward p will be positive with a value 5-4 = 1. We use 1/4p as the value of A in the vertex form for the equation of a parabola: y = A(x - h)² + k (where (h, k) is the vertex.

So, y = 1/4(1) (x + 2)² + 4 = 1/4 (x + 2)²+4.

The following graph also shows that any point on the parabola will be the same distance from the directrix as it is from the focus--that is the definition of a parabola.

desmos.com/calculator/o4jqixzlim

You can visit the above graph and slide point R along the parabola and make note that those distances remain equal regardless of the position of R. If you slide R to coincide with the vertex you will see that the distance to directrix is 1 and the distance to focus is 1, giving an indication as to why the vertex if the midpoint of that segment.

Answer 3

Vertex is half way between focus and directrix.

Verterx: (-2, 2).

f(x) = a(x + 2)² + 2

a = 4p, p = distance betwee focus and vetex

p = 2, 4p = 8

f(x) = 8(x + 2)² + 2