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:
The standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2 is [tex]y=\frac{1}{8}x^2[/tex].
Step-by-step explanation:
The standard form of the parabola is
[tex]y=ax^2+bx+c[/tex]
The general form of the parabola is
[tex](x-h)^2=4p(y-k)[/tex] ..... (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
[tex](h,k+p)=(0,2)[/tex]
[tex]h=0[/tex]
[tex]k+p=2[/tex] .... (2)
[tex]y=k-p\Rightarrow k-p=-2[/tex] .... (3)
On solving (2) and (3), we get
[tex]k=0[/tex]
[tex]p=2[/tex]
Substitute h=0, k=0 and p=2 in equation (1).
[tex](x-0)^2=4(2)(y-0)[/tex]
[tex]x^2=8y[/tex]
Divide both sides by 8.
[tex]\frac{1}{8}x^2=y[/tex]
Therefore the standard form of the equation of the parabola with a focus at (0, 2) and a directrix at y = -2 is [tex]y=\frac{1}{8}x^2[/tex].
