Question

Find the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2.

Answer 1

Since the focus is at (0, 2) and directrix is y = -2

point where both of these have the same x-value will be at (0, 2) for the focus and (0, -2) for the directrix.

The vertex will also have the same x-value so it will be (0, y).

y-value is half-way between the y-value of the focus, and the y-value of the directrix at x = 0.

Directrix y-value is -2 at x = 0 and for the focus it's 2 at x = 0. 

Halfway between y = -2 and y = 2 is y = 0. 

So the vertex of the parabola occurs at (0, 0). 

So that's x^2 = 4ay = 4(2)y = 8y. 

y = 1/8*x^2

hope it helps

Answer 2

Answer:

The standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2 is $y=\frac{1}{8}x^2$.

Step-by-step explanation:

The standard form of the parabola is

$y=ax^2+bx+c$

The general form of the parabola is

$(x-h)^2=4p(y-k)$         ..... (1)

Where, (h,k) is vertex, (h,k+p) is focus and y=k-p is directrix.

It is given that the parabola with a focus at (0, 2) and a directrix at y = -2. It means

$(h,k+p)=(0,2)$

$h=0$

$k+p=2$       .... (2)

$y=k-p\Rightarrow k-p=-2$       .... (3)

On solving (2) and (3), we get

$k=0$

$p=2$

Substitute h=0, k=0 and p=2 in equation (1).

$(x-0)^2=4(2)(y-0)$

$x^2=8y$

Divide both sides by 8.

$\frac{1}{8}x^2=y$

Therefore the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2 is $y=\frac{1}{8}x^2$.