SOLUTION: find the focus and directrix of the parabola y^2=14x and sketch

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the focus and directrix of the parabola y^2=14x and sketch      Log On


   

Question 1087501: find the focus and directrix of the parabola y^2=14x and sketch
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

the parabola y%5E2=14x+ is sideways parabola opens towards positives
x=y%5E2%2F14+ or x=%281%2F14%29y%5E2+
standard formula is: x=a%28y-k%29%5E2%2Bh+ where h and k are coordinates of vertex
in your a=%281%2F14%29,h=0 and k=0
the focus is at
(h%2B1%2F4a, k)
=(%280%2B1%29%2F4%281%2F14%29, 0)
=(14%2F4, 0)
=(7%2F2, 0)
the vertex is exactly midway between the directrix and the focus, so directrix is x=-7%2F2

Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
%28y-k%29%5E2=4p%28x-h%29

has vertex (h,k), and distance from vertex to both focus
and directrix is |p|.  If p is positive the parabola opens
right with the vertical directrix is |p| units left of the 
vertex and the focus is |p| units right of the vertex.  

%28y-0%29%5E2=14%28x-0%29

has vertex (0,0), and distance from vertex to both focus
and directrix is |14/4| or 7/2.  Since 7/2 is positive the 
parabola opens right with the vertical directrix 7/2 units  
left of the vertex and the focus is 7/2 right of the vertex.  

So the focus is 4 units right of vertex (0,0) which is (4,0),
and the dirctrix is a vertical line 7/2 units left of the 
vertex (0,0) which is the vertical line x = -7/2



Edwin