Find the vertex,fcus, ad directrix of the parabola (x+2)to the scond power  = -24(y-1)

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hala718 eNotes educator| Certified Educator

The parabola (x+2)^2 = -24(y-1)

We need to find the vertex, focus, and directions

1. To find the vertex: We know that in the standard parabola x^2 = 4ay , the vertex (x,y)=(0,0), then in (x+2)^2 = 24(y-1). the vertex (x+2, y-1) = (0.0).

==> x+2 = 0 ==> x=-2

and y-1=0 ==> y=1

then the vertex is (x,y)=(-2,1)

2. To find the focus, first we find the focal length a.

First we write the parabola in the standard formula,

(x+2)^2= 4(-6)(y-1)

then a = -6

==> the focus is (x,a) = (-2,-6)

3. The directrix (for the standard parabola, x= -a)

Then, for our parabola  x+2 = -(-6)=6

==> x = 4 is the directrix.


neela | Student

To find the vertex, focus and directrix of the parabola:

(x+2)^2 = -24(y-1)


The standard equation of the parabola is y ^2 = 4ax with x axis as a symmetriacal axis having (0,0) as vertex, (a,o) as the focus, and  a as focal length, and x = -a as directrix.

A similar parobola is  X^2 = 4a *Y...............(1), which is a parabola, symmetrical about y axis with (0,0) as vertex, (0,a) as focus and Y = -a as directrix and  a as focal length.

Now compare  (x+2)^2 = -24(y-1) Or

(x+2)^2 = 4*(-6)(y-1) ... ....(2), the given parabola,  with the standard parabola at eq(1).

We get for (X,Y)  = (0,0),  (x+2, y-1) = (0,0). So  (x,y)  =   (-2,1) is the vertex of the given parabola.

Focal length:  4a = 4*(-6) or a = -6 is the focal length implies the focal length is below the origin  by  6 units on y axis.

The focus:  (X,Y) = (0,a) is the focus of the standard parabola. Corresponding to this,  we get (x+2, y) = (0, -6) Or

(x,y) = (-2,-6) is the coordinate position of the focus  for the given parabola.

The directrix :  X = -a. Or x+2 = -(-6).  Or for the given equation of the parabola, x+2 = 6  is the directrix. Or

x = 6-2. Or

x = 4 is the directrix of the parabola.