Geo5.3.9

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What are the coordinates of the foci of the hyperbola {\frac  {(x+1)^{2}}{25}}-{\frac  {(y+2)^{2}}{16}}=1\,

From the given equation h=-1,k=-2,a=5,b=4\,

eccentricity e={\frac  {{\sqrt  {25+16}}}{5}}={\frac  {{\sqrt  {41}}}{5}}\,

The foci are (h\pm ae,k)=(-1\pm {\sqrt  {41}},-2)\,

(-1+{\sqrt  {41}},-2),(-1-{\sqrt  {41}},-2)\,


Main Page:Geometry:Hyperbola