Question

*Helper will recieve a medal :)*
identify conesection and state important parts
y^2=x+16y

Answer 1

What are considered "important parts"? Is that:
vertex, focus, directrix
any others we need to concern ourselves with?

Answer 2

vertices,co-vertices, foci, radius center depends on what it is?

Answer 3

Oh got it. I had already identified this as a parabola because only the y was squared. There is no squared term for x, so it does not become an ellipse/circle or hyperbola.

Answer 4

Parabolas always have the form: y = ax^2 or x = ay^2 barring any extras such as vertex shifts

Answer 5

alright

Answer 6

So we should solve this equation for x first:
y^2 = x+16y
and then try to put it into vertex form of a parabola by completing the square with y^2 - 16y. You know how to do that part?

Answer 7

would it be y^2-16y+64?

Answer 8

Yup.
y^ 2 - 16y + 64 - 64 = x

Answer 9

then that would be (y-8)^2=x

Answer 10

There will still be the - 64 part because otherwise we have changed the equation' s value
(y^ 2 - 16y + 64) - 64 = x
(y - 8)^2 - 64 = x

Answer 11

now how do you find the vertex, focus and directrix?

Answer 12

Vertex is the easiest to find first. Starting with the equation with the vertex at (0, 0) and applying some transformations to get it to match our current equation...
y^2 = x
We shifted the y-value UP 8 units, which makes for subtracting (it takes greater y-values to reach the same point now)
(y - 8)^2 = x
and then shifted it to the LEFT 64 units:
(y - 8)^2 - 64 = x
So our vertex has moved: (0 - 64, 0 + 8) = (-64, 8)

Answer 13

The focus and directrix of the parabola are subject to the same translations.
x = ay^2 The distance from vertex to directrix or focus is c= 1/4a
The orientation of the parabola puts the focus in front of (horizontally) the vertex, and the directrix vertical behind the parabola: