SOLUTION: how do i classify conic sectins n write its equation in standard form? example: -2y^2+x-20y-49=0

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Question 571627: how do i classify conic sectins n write its equation in standard form?
example:
-2y^2+x-20y-49=0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The conic sections you are likely to encounter will have axes of symmetry parallel to the x and y axes, and that will make it easier, because you will not see a term with xy.
Getting to the standard form involves completing squares. Whenever you see a variable and a variable squared, as in
-2y%5E2-20y=-2%28y%5E2%2B10y%29 you have to imagine the part of the expression with that variable as part of a perfect square.
In this case
%28y%2B5%29%5E2=y%5E2%2B10y%2B25 should come to mind.
With that in mind, you start transforming the equation
-2y%5E2%2Bx-20y-49=0 --> -2y%5E2-20y%2Bx-49=0 --> -2%28y%5E2%2B10y%29%2Bx-49=0 --> -2%28y%5E2%2B10y%2B25-25%29%2Bx-49=0 --> -2%28%28y%2B5%29%5E2-25%29%2Bx-49=0 --> -2%28y%2B5%29%5E2%2B50%2Bx-49=0 --> -2%28y%2B5%29%5E2%2Bx%2B1=0 --> 2%28y%2B5%29%5E2=x%2B1 --> %28y%2B5%29%5E2=%281%2F2%29%28x%2B1%29
The last equation tells you that it is a parabola with horizontal axis of symmetry y=-5, vertex at (-1,-5), and focal distance %281%2F2%29%281%2F4%29=1%2F8
It opens to the right. x%2B1%3E=0 ,--> x%3E=-1.
The directrix is the vertical linex=-1-1%2F8=-9%2F8=-1.125.
The focus has x=-1%2B1%2F8=-7%2F8=-0.875, so it's the point(-0.875,-5).