Question 571627
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^2-20y=-2(y^2+10y)}}} you have to imagine the part of the expression with that variable as part of a perfect square.
In this case
{{{(y+5)^2=y^2+10y+25}}} should come to mind.
With that in mind, you start transforming the equation
{{{-2y^2+x-20y-49=0}}} --> {{{-2y^2-20y+x-49=0}}} --> {{{-2(y^2+10y)+x-49=0}}} --> {{{-2(y^2+10y+25-25)+x-49=0}}} --> {{{-2((y+5)^2-25)+x-49=0}}} --> {{{-2(y+5)^2+50+x-49=0}}} --> {{{-2(y+5)^2+x+1=0}}} --> {{{2(y+5)^2=x+1}}} --> {{{(y+5)^2=(1/2)(x+1)}}}
The last equation tells you that it is a parabola with horizontal axis of symmetry {{{y=-5}}}, vertex at (-1,-5), and focal distance {{{(1/2)(1/4)=1/8}}}
It opens to the right. {{{x+1>=0}}} ,--> {{{x>=-1}}}.
The directrix is the vertical line{{{x=-1-1/8=-9/8=-1.125}}}.
The focus has {{{x=-1+1/8=-7/8=-0.875}}}, so it's the point(-0.875,-5).
{{{graph(300,300,-5,45,-9,1,-5+sqrt(2+2x)/2,-5-sqrt(2+2x)/2)}}}