Question

question 1 (2 points)
(02 08)
derive the equation of the parabola with a focus at (0, 1) and a directrix of y = -1.
o a
$f(x)=-\frac{1}{4}x^{2}$
o b
$f(x)=\frac{1}{4}x^{2}$
o c
$f(x)=-4x^{2}$
o d
$f(x)=4x^{2}$

Explanation:

Step1: Recall the definition of a parabola

The distance from any point $(x,y)$ on the parabola to the focus $(0,1)$ is equal to the distance from the point $(x,y)$ to the directrix $y = - 1$.
The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, and the distance from the point $(x,y)$ to the line $y=-1$ is $|y + 1|$.
The distance from $(x,y)$ to $(0,1)$ is $\sqrt{(x - 0)^2+(y - 1)^2}$.
So, $\sqrt{(x-0)^2+(y - 1)^2}=|y + 1|$.

Step2: Square both sides

Squaring both sides of the equation $\sqrt{x^{2}+(y - 1)^{2}}=|y + 1|$, we get $x^{2}+(y - 1)^{2}=(y + 1)^{2}$.
Expand both sides: $x^{2}+y^{2}-2y + 1=y^{2}+2y+1$.

Step3: Simplify the equation

Subtract $y^{2}+1$ from both sides of the equation $x^{2}+y^{2}-2y + 1=y^{2}+2y+1$.
We have $x^{2}-2y=2y$.
Then, $x^{2}=4y$, or $y=\frac{1}{4}x^{2}$, which can be written as $f(x)=\frac{1}{4}x^{2}$.

Answer:

B. $f(x)=\frac{1}{4}x^{2}$