SOLUTION: Sketch the graph and identify focus, directrix, vertex, latus rectum and axis of symmetry: y^2 = -25x Thank you very much!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Sketch the graph and identify focus, directrix, vertex, latus rectum and axis of symmetry: y^2 = -25x Thank you very much!       Log On


   



Question 1087427: Sketch the graph and identify focus, directrix, vertex, latus rectum and axis of symmetry: y^2 = -25x
Thank you very much!

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
 This is exactly like the other one I did for you, except
it opens left instead of right.  Hey, you've got to learn this 
stuff, because you have to pass the tests all by yourself.  

%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 negative the parabola opens
leftt with the vertical directrix is |p| units right of the 
vertex and the focus is |p| units left of the vertex.  

%28y-0%29%5E2=-25%28x-0%29

has vertex (0,0), and distance from vertex to both focus
and directrix is |-25/4| or 25/4.  Since -25/4 is negative the 
parabola opens left with the vertical directrix 25/4 units  
right of the vertex and the focus is 25/4 left of the vertex.  

So the focus is 25/4 units left of vertex (0,0) which is (-25/4,0),
and the directrix is a vertical line 25/4 units right of the 
vertex (0,0) which is the vertical line x = 25/4



Edwin