Asked by Tyler
                I have a question and I can not figure it out. Any answers to the following along with any explanation would be great!
Here is the equation:
x=.2y^2-2y-3
Now I know that is an equation for a parabola. Now, I need the following:
Equation in standard form
Vertices
Co-vertices
Center
Foci
Directric
Major Axis
Minor Axis
Some may not be needed based on it being a parabola. Any help would be great!
            
        Here is the equation:
x=.2y^2-2y-3
Now I know that is an equation for a parabola. Now, I need the following:
Equation in standard form
Vertices
Co-vertices
Center
Foci
Directric
Major Axis
Minor Axis
Some may not be needed based on it being a parabola. Any help would be great!
Answers
                    Answered by
            Damon
            
    Here is the equation:
x=.2y^2-2y-3
x+3 = .2 y^2 -2 y
multiply both sides by 5 to get 1 as coefficient of y^2
5 x + 15 = y^2 - 20 y
add 100 to both sides to get perfect square on the right
5 x + 115 = y^2 -20 y +100 = (y-10)^2
so
(y-10)^2 = 5 ( x + 23)
vertex at (-23,10) etc
    
x=.2y^2-2y-3
x+3 = .2 y^2 -2 y
multiply both sides by 5 to get 1 as coefficient of y^2
5 x + 15 = y^2 - 20 y
add 100 to both sides to get perfect square on the right
5 x + 115 = y^2 -20 y +100 = (y-10)^2
so
(y-10)^2 = 5 ( x + 23)
vertex at (-23,10) etc
                    Answered by
            Tyler
            
    Wow thanks for that. Could you help me with all the other things I listed above. This stuff is confusing to me.
    
                    Answered by
            Damon
            
    Here is the equation:
x=.2y^2-2y-3
x+3 = .2 y^2 -2 y
multiply both sides by 5 to get 1 as coefficient of y^2
5 x + 15 = y^2 - 10 y
add 25 to both sides to get perfect square on the right
5 x + 40 = y^2 - 10 y + 25 = (y-5)^2
so
(y-5)^2 = 5 ( x + 8)
vertex at (-8,5) etc
    
x=.2y^2-2y-3
x+3 = .2 y^2 -2 y
multiply both sides by 5 to get 1 as coefficient of y^2
5 x + 15 = y^2 - 10 y
add 25 to both sides to get perfect square on the right
5 x + 40 = y^2 - 10 y + 25 = (y-5)^2
so
(y-5)^2 = 5 ( x + 8)
vertex at (-8,5) etc
                    Answered by
            Tyler
            
    Actually, I belive you made a mistake there. Multiplying both sides by five would get my -10y and therefor, I would add 25 to both sides right?
    
                    Answered by
            Damon
            
    (y-5)^2 = 5 ( x + 8) 
y gets big + as x gets big + so opens to the right
well, all that is left is a focus and directrix. parabolas do not have major and minor axes like ellipses.
if in the form
(y-k)^2 = 4 a (x-h)
so here a = 5/4
then
vertex at (h,k)
vertex to focus = a so at x= -8+5/4 = -6.75 so (-6.75 , 5)
vertex to directrix = a so at x = -9.25
    
y gets big + as x gets big + so opens to the right
well, all that is left is a focus and directrix. parabolas do not have major and minor axes like ellipses.
if in the form
(y-k)^2 = 4 a (x-h)
so here a = 5/4
then
vertex at (h,k)
vertex to focus = a so at x= -8+5/4 = -6.75 so (-6.75 , 5)
vertex to directrix = a so at x = -9.25
                    Answered by
            Tyler
            
    When you say vertex, you mean the center so what about vertices and co-vertices. And there is only 1 foci?
    
                    Answered by
            Tyler
            
    Another thing, would a=1.25?  Because 4*1.25 equals 5
    
                    Answered by
            Damon
            
    yes, 1.25 = 5/4
when I say vertex I mean the leftmost point of the parabola. It has no center like an ellipse and it only has one vertex and one focus.
Look in your text book for labeled pictures of:
parabola
ellipse
hyperbola
    
when I say vertex I mean the leftmost point of the parabola. It has no center like an ellipse and it only has one vertex and one focus.
Look in your text book for labeled pictures of:
parabola
ellipse
hyperbola
                    Answered by
            Tyler
            
    Sorry but I am trying to understand your reasoning for the directrix. How did you get -9.25. I would think it would be -1.25.
    
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