Given a parabola with a focus of (-2, -4) and a directrix of y=2 - write the quadratic equation that describes the parabola

Question

We are making an equation with the graph

Answer 1

The important parameter for a parabola is the P value.   P is the distance from the vertex to the directrix.

In this problem the vertex is below the horizontal directrix by  2 - (-4) = 6.    So P =6.

 

In terms of the vertex  formula  y = a (x - (-2))² +k        P = 1/4|a|  .

The a value will be negative because this parabola opens downward away from the directrix

This implies that a = -1/24.   

The formula so far is thus  y = - (1/24) (x + 2)²  + k.  

However  y(-2) = -4  so  k = -4

 

Final result in vertex form is  y =  -(1/24) ( x +2)² -4 

In standard form this is   -(1/24) x² - (1/6) x - 25/6

 

The key formula is the one above for Pm in terms of a

 

 

Answer 2

Molly,

 

The canonical form of a parabola with a vertex at (h,k) is (x - h)² = 4p(y - k).

 

The focus (-2, -4) and the vertex (h, k) are always along the same x-coordinate, so h = -2.

 

The focus (-2, -4) is in the third quadrant and is below the directrix y = 2, so the parabola is inverted, and the sign of p is negative.

 

The distance between the focus (-2, -4) and the directrix y = 2 along the x-coordinate x = -2 is |p| = 2 - (-4) = 6.

 

The vertex is half way 6/2 between the focus and the directrix at y = 2, so it's y-coordinate is  k = y = 2 - 6/2 = -1.

 

The vertex is at (h, k) = (-2, -1).

 

Substituting the parameters into the canonical equation: (x - (-2))² = 4(-6)(y - (-1)) => (x + 2)² = -24(y+1)

 

Michael.