SOLUTION: Find the foci for the hyperbola, using this equation (x+2)^2/81 - (y+4)^2/4 = 1.It tells me to put it in this form (x1,y1),(x2,y2).Please help I don't know how to solve this.
Algebra ->
Quadratic-relations-and-conic-sections
-> SOLUTION: Find the foci for the hyperbola, using this equation (x+2)^2/81 - (y+4)^2/4 = 1.It tells me to put it in this form (x1,y1),(x2,y2).Please help I don't know how to solve this.
Log On
Question 401170: Find the foci for the hyperbola, using this equation (x+2)^2/81 - (y+4)^2/4 = 1.It tells me to put it in this form (x1,y1),(x2,y2).Please help I don't know how to solve this. Answer by ewatrrr(24785) (Show Source):
Hi
Standard Form of an Equation of an Hyperbola is where Pt(h,k) is a center,
and the vertices are 'a' units right and left of center.
(x+2)^2/81 - (y+4)^2/4 = 1 | a = 9 for this hyperbola
Center is Pt(-2,-4), therefore the vertices would be (-11,-4) and (7,-4)
Foci: c = sqrt(a^2 + b^2)}}} = sqrt(85)= 9.22
Pt(-11.22, -4) and Pt(7.22, -4)
