Question

find the equation of the parabola with the given focus and directrix.
focus $(-2,3)$, directrix $y=-2$
which equation below is of a parabola with the given focus and directrix?
$\bigcirc$ a. $y = 10(x + 2)^2 - 1$ $\bigcirc$ b. $y = \frac{1}{10}(x - 2)^2 + \frac{1}{2}$
$\bigcirc$ c. $y = \frac{1}{10}(x + 2)^2 - \frac{1}{2}$ $\bigcirc$ d. $y = 10(x - 2)^2 + 1$
$\bigcirc$ e. $y = 10(x + 2)^2 + 1$ $\bigcirc$ f. $y = \frac{1}{10}(x - 2)^2 - \frac{1}{2}$
$\bigcirc$ g. $y = \frac{1}{10}(x + 2)^2 + \frac{1}{2}$ $\bigcirc$ h. $y = 10(x - 2)^2 - 1$

Explanation:

Step1: Find the vertex of the parabola

The vertex \((h,k)\) of a parabola is the mid - point between the focus \((x_f,y_f)\) and the directrix \(y = y_d\).
For a focus \((-2,3)\) and directrix \(y=-2\), the \(x\) - coordinate of the vertex \(h=-2\) (since the parabola is vertical, the \(x\) - coordinate of the vertex is the same as the \(x\) - coordinate of the focus).
The \(y\) - coordinate of the vertex \(k=\frac{y_f + y_d}{2}=\frac{3+( - 2)}{2}=\frac{1}{2}\).

Step2: Find the value of \(a\)

The distance \(a\) between the vertex \((h,k)\) and the focus (or the vertex and the directrix) for a parabola of the form \(y=a(x - h)^2+k\) is given by \(a=\frac{1}{4d}\), where \(d\) is the distance between the focus and the vertex (or the vertex and the directrix).
The distance \(d\) between the focus \((-2,3)\) and the vertex \((-2,\frac{1}{2})\) is \(d = 3-\frac{1}{2}=\frac{5}{2}\). Then \(a=\frac{1}{4\times\frac{5}{2}}=\frac{1}{10}\)

Step3: Write the equation of the parabola

The standard form of a vertical parabola is \(y=a(x - h)^2+k\). Substituting \(h=-2\), \(k = \frac{1}{2}\) and \(a=\frac{1}{10}\) into the equation, we get \(y=\frac{1}{10}(x + 2)^2+\frac{1}{2}\)

Answer:

G. \(y=\frac{1}{10}(x + 2)^2+\frac{1}{2}\)