9x^2+54x-y^2+10y+46=0 Find the center vertices foci and assymptotes

Question

Find the center foci vertices assymptotes for this Hyperbola

Answer 1

9x²+54x-y²+10y+46=0.

To find the centre of the hyperbola we need to complete two squares:

9x²+54x-y²+10y+46=9(x²+6x)-(y²-10y)+46=0,

9(x²+6x+9-9)-(y²-10y+25-25)+46=0,

9(x+3)²-81-(y-5)²+25+46=0,

9(x+3)²-(y-5)²=10 has centre (-3,5).

The asymptotes are found from 9(x+3)²-(y-5)²=0, that is:

9(x+3)²=(y-5)², 3(x+3)=±(y-5).

3x+9=y-5⇒y=3x+14; 3x+9=-y+5⇒y=-3x-4. These are the equations of two intersecting lines and are the asymptotes.

The foci and the centre of the hyperbola are collinear, defining its horizontal axis. Each focus is the same distance from the centre of the hyperbola. It's easier to redefine the hyperbola as if the centre is at the origin (0,0) so that we have a simpler equation:

x²/a²-y²/b²=1, which would be 0.9x²-0.1y²=1, making a²=1/0.9=10/9 and b²=1/0.1=10, so a=√10/3 and b=√10.

Eccentricity, e=√(1+b²/a²)=√(1+9)=√10 and the focal length c=ae=√(a²+b²)=√(10/9+10)=10/3, the distance between the hyperbola's centre and each focus. The foci lie on the line y=5. The x-coordinates of the foci are -3-10/3=-19/3 and -3+10/3=⅓, so the foci are at (-19/3,5) and (⅓,5).