Question

Derive the equation of the parabola with a focus at (0, 1) and a directrix of y = −1. (2 points)

Answer 1

To determine the equation of the parabola with the given features above, we need to understand or know what are the significance of those values. The directrix is a line that is perpendicular to the axis of symmetry of a curve. A parabola is defined by its focus and the directrix. The parabola is a group of points where the distance to its focus is equal to the distance to its directrix. So, the directrix for this case is all points (x, -1). By using the distance formula, we construct the equation.

√((x-0)^2 + (y-1)^2) = √((x-x)^2 + (y+1)^2)
x^2 + (y-1)^2 = (y+1)^2
x^2 + y^2 - 2y+1 = y^2 +2y + 1
x^2 - 2y = 2y
x^2 = 4y
y = x^2/4

Answer 2

Answer:

$y=4(x)^2$

Step-by-step explanation:

 Focus =(h,k+p)  

Directrix= y=k-p

Focus given is : (0,1)

And directrix given is : y=-1

(h,k+p) =(0,1)

On comparing the values we get

h=0 and k+p=1

y=k-p= -1

Hence, we gave two equations

k+p=1  and k-p= -1

Substitute p= 1-k in k-p= -1 we get:

k-(1-k) = -1

k-1+k= -1

k=0

Now, we get

h=0 and k=0

And put k=0 in p = 1-k

We get p=1

We have general equation of parabola

$(y-k)=4p(x-h)^2$

$(y-0)=4(1)(x-0)^2$

$y=4(x)^2$

Hence, the required equation is :

$y=4(x)^2$