SOLUTION: Find the vertex,focus and end points of the latus rectum of the parabola y^2-6x+3=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex,focus and end points of the latus rectum of the parabola y^2-6x+3=0      Log On


   

Question 1018492: Find the vertex,focus and end points of the latus rectum of the parabola y^2-6x+3=0
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Knowing something about how an equation for parabola is derived is helpful.

An easy transformation of the equation is 6%28x-1%2F2%29=y%5E2. Your parabola has a vertex at the left and parabola opens toward the right. The vertex is (1/2,0).

Comparing to a more generally derived equation, you could have 4p%28x-1%2F2%29=y%5E2 for p being distance from vertex to focus, so here, you have p=3%2F2. You will expect the focus to be to the right of the vertex, for your example, and so the focus point is (2,0).

(Latus Rectum is related to where the y values are in relation to the focus.)
(What are the y values for x=2 ?)