Question

How do you find the vertex, focus and directrix of `y^2+10y-20x+37=0`?

Answer 1

The vertex is `V=(3/5, -5)`
The focus is `F=(28/5,-5)`
The directrix is `x=-22/5`

Explanation:

Let's rearrange the equation and complete the squares

`y^2+10y-20x+37=0`

`y^2+10y=20x-37`

`y^2+10y+25=20x-37+25`

`(y+5)^2=20x-12=20(x-12/20)=20(x-3/5)`

We compare this equation to the standard equation of a parabola

`(y-b)^2=2p(x-a)`

`p=10`

The vertex is `V=(a,b)=(3/5, -5)`

The focus is `F=(a+p/2, b)=(28/5,-5)`

The directrix is `x=a-p/2=3/5-5=-22/5`

graph{(y^2+10y-20x+37)(y+2200x/5+2000)((x-3/5)^2+(y+5)^2-0.01)((x-28/5)^2+(y+5)^2-0.01)=0 [-10, 10, -5, 5]}