Question

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

Answer 1

Answer: -(x-2)^2-3/4

Answer 2

keeping in mind that, there's a "p" distance from the vertex to the focus, and the same distance from the vertex to the directrix. Therefore then, the vertex is half-way between those two fellows.

now, we know the focus point is at 2, -1, and the directrix is below it at y = -½, therefore, is a vertical parabola, opening upwards.

check the picture below.

from y = -1 to y = -½, there's only ½ units, and the vertex is right in the middle, and half of ½ is ¼, then the y-coordinate for the vertex must be at -1 - ¼, or -5/4 then.

since the parabola is opening upwards, the "p" unit is positive, ¼.


`\bf \textit{parabola vertex form with focus point distance}\\\\ \begin{array}{llll} (y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\ \boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}})} \end{array} \qquad \begin{array}{llll} vertex\ ({{ h}},{{ k}})\\\\ {{ p}}=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array}\\\\ -------------------------------`


`\bf \begin{cases} h=2\\ k=-(5)/(4)\\ p=(1)/(4) \end{cases}\implies (x-2)^2=4\left( (1)/(4) \right)\left[y - \left(-(5)/(4) \right) \right] \\\\\\ (x-2)^2=1\left( y+(5)/(4) \right)\implies (x-2)^2= y+(5)/(4)\implies (x-2)^2-(5)/(4)= y`