Answer:
Hyperbola
Step-by-step explanation:
[tex]x^2-4x-y^2+6y-9=5[/tex]
Lets write variables in square form using completing the square method
LEts group x and y terms
[tex](x^2-4x)+(-y^2+6y)-9=5[/tex]
y^2 terms should be positive always, so we pull out negative sign from y^2
[tex](x^2-4x)-(y^2-6y)-9=5[/tex]
We have negative sign in the middle. so we will get equation in the form of
[tex]\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} =1[/tex]
If we have negative sign in the middle then it is an hyperbola
