Answer
Vertex: $(0,0)$
Focus: $\left(0,-\dfrac{E}{4A}\right)$
Directrix: $y=\dfrac{E}{4A}$
Work Step by Step
We are given the equation:
$Ax^2+Ey=0$, $A\not=0,E\not=0$
Rewrite the equation:
$Ax^2=-Ey$
$x^2=-\dfrac{E}{A}y$
The equation is in the form:
$(x-h)^2=4p(y-k)$
Therefore the equation is that of a parabola. Because the equation contains $x^2$, the parabola is vertical.
Determine $h,k,p$:
$h=0$
$k=0$
$4p=-\dfrac{E}{A}\Rightarrow p=-\dfrac{E}{4A}$
Determine the vertex:
$(h,k)=(0,0)$
Determine the focus:
$(h, k+p)=\left(0,0-\dfrac{E}{4A}\right)=\left(0,-\dfrac{E}{4A}\right)$
Determine the directrix:
$y=h-a$
$y=0-\left(-\dfrac{E}{4A}\right)$
$y=\dfrac{E}{4A}$
We are given the equation:
$Ax^2+Ey=0$, $A\not=0,E\not=0$
Rewrite the equation:
$Ax^2=-Ey$
$x^2=-\dfrac{E}{A}y$
The equation is in the form:
$(x-h)^2=4p(y-k)$
Therefore, the equation is that of a parabola. Because the equation contains $x^2$, the parabola is vertical, and its axis is vertical.
Determine $h,k,p$:
$h=0$
$k=0$
$4p=-\dfrac{E}{A}\Rightarrow p=-\dfrac{E}{4A}$
Determine the vertex:
$(h,k)=(0,0)$
Determine the focus:
$(h, k+p)=\left(0,0-\dfrac{E}{4A}\right)=\left(0,-\dfrac{E}{4A}\right)$
Determine the directrix:
$y=h-a$
$y=0-\left(-\dfrac{E}{4A}\right)$
$y=\dfrac{E}{4A}$
