SOLUTION: Graph the equation. Identify the center, vertices and Foci. {{{ (x+3)^2/16 - (y-1)^2/25=1 }}}

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Graph the equation. Identify the center, vertices and Foci. {{{ (x+3)^2/16 - (y-1)^2/25=1 }}}      Log On


   

Question 625961: Graph the equation. Identify the center, vertices and Foci.
+%28x%2B3%29%5E2%2F16+-+%28y-1%29%5E2%2F25=1+
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Graph the equation. Identify the center, vertices and Foci
%28x%2B3%29%5E2%2F16+-+%28y-1%29%5E2%2F25=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation:%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1
For given equation:
center: (-3,1)
a^2=16
a=√16=4
vertices: (-3±a,1)=(-3±4,1)=(-7,1) and (1,1)
..
b^2=25
b=√25=5
..
c^2=a^2+b^2=16+25=41
c=√41≈6.4
Foci: (-3±c,1)=(-3±6.4,1)=(-9.4,1) and (3.4,1)
see graph below:
y=±(25(x+3)^2/16-25)^.5+1