How do you find the center and vertex for #9x^2-16y^2+18x+160y-247=0#?
This represents a hyperbola. The center is at (-1, 5). The vertcies are #(-1, 9) and (-1, 10)
This is the most general method for any second degree equation.
There is no xy-term and the product of the coefficients of #x^2 and
y^2 = -144<0#. So this equation represents a hyperbola..
The equation has the form
Comparing coefficients in the expansion with the given coefficients,
Now, the standard form of the given equation is
The asymptotes of this hyperbola are given by
So, the center of the hyperbola is the point of intersection C(-1, 5)..
Shifting the origin to the center using X = x+1 and Y = y-5, the equation becomes
The semi-axes are
The major axis is Y-axis (x=-1) and the vertices are X= 0, Y=+-A=+-3.
Referred to given axes,
The two vertices are
Foci are on the major axis and are distant Ae = 5, from the center C(-1, 5).
So, the two foci are
Just to give the most general method, I have avoided looking for other easier methods, for particular cases..