Question

Find the coordinates of the focus and the equation of the directrix for 2x²+5y=0?

Answer 1

Focus `(0,-1/10)` and directrix `y=1/10.`

Explanation:

`2x^2 + 5y = 0`

`y = -2/5 x^2`

I remember the focus and directrix of `y=x^2` are `(0,1/4)` and `y=-1/4.` Generalizing slightly, `y=ax^2` has focus `(0, a/4)` and directrix `y= -a/4.`

Here we have `a=-2/5` for focus `(0,-1/10)` and directrix `y=1/10.`

Let's plot these. Here's the tricky thing I enter to plot multiple functions and points with the Socratic `cancel{text{crasher}}` grapher.

0=(y-1/10)(2x^2+5y)( x^2+ (y+1/10)^2 - .02^2)

graph{0=(y-1/10)(2x^2+5y)( x^2+ (y+1/10)^2 - .01^2) [-1.25, 1.25, -0.625, 0.625]}