Question

Find the equation for a parabola with vertex (-2, 0) and point on the graph (0, 1) [drawing below]
I assume that the focus is (0, 0), so a = 2, so putting all these things into an equation I get:
(y-0)^2 = 4(2)[x+2]
y^2 = 8(x+2)
But the book says it's
y^2 = (1/2)(x+2)

What am I doing wrong?

Answer 1

Here's the graph shown in the book:

Answer 2

The latus rectum is a segment so I'm wondering why you are giving it as if it were the point (0,1).

Answer 3

Oh, sorry, I meant that it's a point on the graph in the same line as the focus. I don't the name for it, but I drew the graph from the book ^

Answer 4

I've been working on it some more and I just realized the book never said that point is in line with the focus.
But now when I try to solve for the focus, it sill doesn't work...
y^2 = 4ax
1^2 = 4a(0)

That gives me nothing!

Answer 5

But the book says it's y^2 = (1/2)(x+2)
---------------------------------

General equation for horizontal parabola

4p ( x - h) = (y- k)²

The vertex (h,k) is given to be (-2,0)

Substitute that into the general formula.

4p ( x - (-2) ) = (y - 0)²

4p (x + 2) = y²

(0,1) is a given point on the parabola

Let (x1,y1) = (0,1)

So, 4p ( (0) + 2 ) = ( (1) - 0) )²

4p(2) = 1

8p = 1

p = (1/8)

----

Returning to this earlier point:

4p (x + 2) = y²

4*(1/8) (x + 2) = y²

(1/2) (x + 2) = y²

y² = (1/2) (x + 2) @MGR Check this solution and see if you agree. Thanks.