## College Algebra (6th Edition)

Foci: $(-2, -1)$ and $(6, -1)$.
\includegraphics[width=159.45mm,height=12.7mm]{TEMP0000.eps} On the LHS, gather all terms containing x: $9x^{2}-36x=$ $=9(x^{2}-4x)$=$9(x^{2}-4x+4-4)$ $9(x-2)^{2}-36$ ...gather the terms containing y: $25y^{2}+50y=$ $=25(y^{2}+2y)=25(y^{2}+2y+1-1)$ $=25(y+1)^{2}-25$ Rewrite the equation, transferring the constants to the RHS: $9(x-2)^{2}+25(y+1)^{2}=164+36+25$ $9(x-2)^{2}+25(y+1)^{2}=225\qquad/\div 225$ $\displaystyle \frac{(x-2)^{2}}{25}+\frac{(y+1)^{2}}{9}=1$ $\displaystyle \frac{(x-2)^{2}}{5^{2}}+\frac{(y+1)^{2}}{3^{2}}=1$ major axis horizontal, center at (2,-1), $a=5,\ b=3$ $c^{2}=a^{2}-b^{2}=25-9=16$ $c=4$ Foci are $c$ units right and $c$ units left of center. Foci: $(-2, -1)$ and $(6, -1)$.