SOLUTION: Find the Vertex, directrix, focus, length of latus rectum ,center , equation /sketch of a parabola ( x - 1 )2 = 2 ( 𝑦+ 2 )

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the Vertex, directrix, focus, length of latus rectum ,center , equation /sketch of a parabola ( x - 1 )2 = 2 ( 𝑦+ 2 )       Log On


   

Question 1178191: Find the Vertex, directrix, focus, length of latus rectum ,center , equation /sketch of a parabola ( x - 1 )2 = 2 ( 𝑦+ 2 )
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Use "^" to denote exponentiation: "(x-1)^2" instead of "(x-1)2".

%28x-1%29%5E2+=+2%28y%2B2%29

The x is squared, so the parabola opens up or down. The vertex form of the equation I prefer to use is

y-k+=+%281%2F%284p%29%29%28x-h%29%5E2

In this form, the vertex is (h,k); and p is the directed distance (i.e., can be negative) from the directrix to the vertex and from the vertex to the focus. With that form of the equation, 4p is the length of the latus rectum.

Put the given equation in that form:

%28x-1%29%5E2+=+2%28y%2B2%29
2%28y%2B2%29+=+%28x-1%29%5E2
y%2B2+=+%281%2F2%29%28x-1%29%5E2

y-%28-2%29+=+%281%2F%284%281%2F2%29%29%29%28x-%281%29%29%5E2

That gives us (h,k)=(1,-2) and p=1/2.

Use the vertex and the value of p to find the focus and the directrix.

graph%28400%2C400%2C-5%2C5%2C-10%2C10%2C%281%2F2%29%28x-1%29%5E2-2%29