SOLUTION: Find the vertex, directrix,focus and the length of the latus rectan of the parabola y^2+4y+3x-4=0.sketch the graph.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, directrix,focus and the length of the latus rectan of the parabola y^2+4y+3x-4=0.sketch the graph.      Log On


   

Question 1024401: Find the vertex, directrix,focus and the length of the latus rectan of the parabola y^2+4y+3x-4=0.sketch the graph.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
y^2+4y+3x-4=0
3x=-y%5E2-4y%2B4, somewhat easy because right side contains a perfect square...
3x=-1%28y%5E2%2B4y-4%29
-3x=y%5E2%2B4y-4
-3x=y%5E2%2B4y%2B4-4-4----completing the square using 4.
-3x=%28y%2B2%29%5E2-8
-3x%2B8=%28y%2B2%29%5E2
highlight%28-3%28x-8%2F3%29=%28y%2B2%29%5E2%29
OR
highlight%283%28x-8%2F3%29=-1%2A%28y%2B2%29%5E2%29

You might want to know why I put the equation into that form, and not into "standard form".

Study this video lesson about equations of parabolas!
Deriving equation for a parabola if given directrix and focus and if vertex is not at Origin

Recognize that YOUR parabola has horizontal symmetry axis, and opens to the LEFT.