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Mathematics 16 Online
OpenStudy (anonymous):

determine the vertices, asymptotes and foci of the hyperbola: 9(x+2)^(2)/25-(y+2)^(2)/9=1

OpenStudy (anonymous):

\[ \frac{9(x+2)^{2}}{25}-\frac{(y+2)^2}{9}=1\\ \frac{(x+2)^{2}}{\frac{25}{9}}-\frac{(y+2)^2}{9}=1\\ a=\frac 53\\ b= 3\\ c^2 = \frac {25}9 + 9\\ c^2= \frac{25+81}9\\ c=\frac {\sqrt{106}}3 \] What are the foci?

OpenStudy (anonymous):

The vertices are on y=-2. \[ \left( -2\pm \frac 5 3, -2 \right) \]

OpenStudy (anonymous):

The foci are \[ \left(-2\pm\frac{\sqrt{106}}{3},-2 \right) \]

OpenStudy (anonymous):

The red point is the center. The Blue points are the vertices. The green points are the foci. The lines are the two asymptotes \[ y=\pm \frac 9 5( x+2) -2\\ \]

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