Question

derive the equation of a parabola with a focus at (-1,0) and a directrix at y = -3. enter your answers in the boxes. be sure to include the correct signs. y = □(x □)^2□

Explanation:

Step1: Find the vertex

The vertex of a parabola is the mid - point between the focus and the directrix. The x - coordinate of the vertex is the same as the x - coordinate of the focus, so $x=-1$. The y - coordinate of the vertex $h$ is $\frac{0+( - 3)}{2}=-\frac{3}{2}$. So the vertex $(h,k)=(-1,-\frac{3}{2})$.

Step2: Determine the value of $p$

The distance $p$ between the vertex and the focus (or the vertex and the directrix) is $|0-(-\frac{3}{2})|=\frac{3}{2}$. Since the focus is above the directrix, the parabola opens upwards and $p=\frac{3}{2}$.

Step3: Use the standard form of the parabola equation

The standard form of a parabola that opens up or down is $y=a(x - h)^2+k$, where $a=\frac{1}{4p}$. Substituting $p = \frac{3}{2}$, we get $a=\frac{1}{4\times\frac{3}{2}}=\frac{1}{6}$.
Substituting $h=-1,k =-\frac{3}{2}$ and $a=\frac{1}{6}$ into the equation $y=a(x - h)^2+k$, we have $y=\frac{1}{6}(x + 1)^2-\frac{3}{2}$.

Answer:

$y=\frac{1}{6}(x + 1)^2-\frac{3}{2}$