Question

Show that an equation of the form $A{x}^2+Ey=0$is the equation of a parabola with vertex at $\left(0,0\right)$and axis of symmetry the y-axis. Find its focus and directrix.

Short answer

The focus is $\left(0,\frac{-E}{4A}\right)$, and equation of directrix is$y=\frac{E}{4A}$.

Step-by-step solution

Step 1. Given information .

Consider the given equation and vertex .

Step 2. Standard form of parabola used .

The standard form of parabola is${\left(x-h\right)}^2=4a\left(y-k\right)$.

Step 3. Find focus and directrix.

To find the focus and directrix rewrite the given equation in the standard form.

${\left(x-0\right)}^2=4\left(\frac{-E}{4A}\right)\left(y-0\right)$

Compare this equation with standard form .

$a=\frac{-E}{4A}$

Therefore the focus is$\left(0,0+a\right)=\left(0,\frac{-E}{4A}\right)$and the line of directrix is$y=h-a=0-\frac{-E}{4A}=\frac{E}{4A}$

Step 4. Plot the graph .

The graph of the equation shown below .