Question

What is the equation of the parabola with a vertex at (-2, 3) and a focus at (-2, 0)? Does it open upward or downward?
a.
(x + 2)883-12-01-00-00_files/i0060000.jpg = 0.4(y - 3); upward
c.
(x + 2)883-12-01-00-00_files/i0060001.jpg = -12(y - 3); downward
b.
(x + 2)883-12-01-00-00_files/i0060002.jpg = 8(y - 3); upward
d.
(x + 2)883-12-01-00-00_files/i0060003.jpg = -8(y - 3); downward

Answer 1

Refer to the diagram shown below.

The distance, d, from the focus to a point P (x,y) is equal to the distance from the directrix to P.
Therefore
d² = (6 - y)²
     = (x + 2)² + y²
That is,
36 - 12y + y² = (x + 2)² + y²
12y = -(x + 2)² + 36

The equation of the parabola is
y = -(1/12)(x + 2)² + 3
It may be written as
-12(y - 3) = (x + 2)²

The leading coefficient -1/12 is negative, therefore the curve opens downward.

Answer:
The equation is (x + 2)² = -12(y - 3).
The curve opens downward.

Answer 2

Answer:

$-12(y-3)=(x+2)^2$ , downward

Step-by-step explanation:

Given : (h,k) = vertex location = (-2, 3)

             Focus = (-2, 0)

This the downward parabola because the vertex is above the focus.

$y-k= -\frac{1}{4p}(x-h)^2$  -----1

where(h,k)is the vertex

p is distance vertex to focus

To Find Distance between focus and vertex we will use distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$(x_1,y_1)=(-2,3)$

$(x_2,y_2)=(-2,0)$

Substitute the values

$d=\sqrt{(-2+2)^2+(0-3)^2}$

$d=\sqrt{3^2}$

$d=3$

So, value of p = 3

Now substitute the values in 1

$y-3=\frac{1}{4(3)}(x+2)^2$

$y-3=\frac{1}{12}(x+2)^2$

$y-3=-\frac{1}{12}(x+2)^2$

$-12(y-3)=(x+2)^2$

Thus the equation of the parabola is $-12(y-3)=(x+2)^2$

Thus Option C is correct: $-12(y-3)=(x+2)^2$ , downward