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22 write an equation of a parabola that opens upward, has a vertex at t…

Question

22 write an equation of a parabola that opens upward, has a vertex at the origin, and a focus at (0, 2)
options:
$x = -\frac{1}{16}y^2$
$x = \frac{1}{16}y^2$
$y = \frac{1}{8}x^2$
$y = -\frac{1}{8}x^2$

Explanation:

Step1: Recall parabola standard form

Parabola opening upward with vertex at origin has form $y=\frac{1}{4p}x^2$, where $p$ is the distance from vertex to focus.

Step2: Identify value of $p$

Focus is at $(0,2)$, so $p=2$.

Step3: Calculate coefficient

Substitute $p=2$ into the formula: $\frac{1}{4p}=\frac{1}{4\times2}=\frac{1}{8}$

Step4: Write final equation

Substitute coefficient into the standard form: $y=\frac{1}{8}x^2$

Answer:

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