Question

Find the standard form of the parabola
with vertex: (- 2,5) and directrix: x = 3.

Answer 1

Check the picture below, so the parabola looks more or less like so, the "p" distance is negative, because the horizontal parabola is opening to the left-hand-side, so

`\textit{horizontal parabola vertex form with focus point distance} \\\\ 4p(x- h)=(y- k)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h+p,k)}\qquad \stackrel{directrix}{x=h-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{`

`\begin{cases} h=-2\\ k=5\\ p=-5 \end{cases}\implies 4(-5)( ~~ x-(-2) ~~ )=(y-5)^2 \implies -20(x+2)=(y-5)^2 \\\\\\ x+2=-\cfrac{1}{20}(y-5)^2\implies {\Large \begin{array}{llll} x=-\cfrac{1}{20}(y-5)^2-2 \end{array}}`