How do you find all the critical points to graph #4x^2 + 20x - 4y^2 + 6y - 3 = 0# including vertices, foci and asymptotes?

1 Answer
Oct 18, 2016

The graph is a hyperbola and we have to write the equation in the standard form.
#4x^2+20x-4y^2+6y-3=4(x^2+5x+25/4)-4(y^2-3/2y+9/16)-3-25-9/4=0#
So the equation becomes
#4(x+5/2)^2-4(y-3/4)^2=121/4#
#(x+5/2)^2/(121/16)-(y-3/4)^2/(121/16)=1#
So the center is #(-5/2,3/4)#
The vertices are #(-5/2+11/4,3/4)# and #(-5/2-11/4,3/4)# which is #(1/4,3/4)and (-21/4,3/4)#
The slope of the asymptotes are 1 and -1
The equations of the asymptotes are #y=3/4+x+5/2# and #y=3/4-x-5/2# that is #y=x+13/4# and #y=-x-7/4#
In order to determine the foci we need #c=+-sqrt(2*(121/6))=+-11/6sqrt2#
So the foci are #(-5/2+)# and ##

graph{4x^2+20x-4y^2+6y-3=0 [-20, 20, -10, 10]}