Question 598932
It is a hyperbola with axes parallel to the x- and y-axes.
{{{y^2 - 16x^2 + 14y + 64x - 31 = 0}}} ---> {{{(y^2 + 14y) - 16(x^2 -4x) = 31}}}
would be a good start.
Then you would complete the squares:
{{{(y^2+14y+49) - 16(x^2-4x+4) = 31+49-16*4}}} --> {{{(y+7)^2-16(x-2)^2=16}}}
Dividing everything by 16, we get
{{{(y+7)^2/16-16(x-2)^2/16=1}}} --> {{{(y+7)^2/4^2-(x-2)^2/1^2=1}}}
So the hyperbola is centered at (2,-7).
The transverse axis is vertical {{{x=2}}}.
When {{{x=2}}} --> {{{(y+7)^2=4^2}}}
The hyperbola crosses the transverse axis at points (2,5) and (2,-19)
with y=4-7=-3 and y=-4-7=-11
The asymptotes cross at center (2,-7), and have the equations
{{{y+7=4(x-2)}}} and {{{y+7=-4(x-2)}}}
that derive from {{{(y+7)^2/12^2=(x-2)^2/3^2}}}
Here is half of that hyperbola, with the asymptotes:
{{{graph(300,300,-3,7,-8,12,4sqrt(1+1(x-2)^2)-7,-4x+1,4x-15)}}}