For the hyperbola $\frac{(x+2)^{2}}{4}-\frac{(y-5)^{2}}{16}=1$
find the center, foci, endpoints of the transverse axis, and the equations of the asymptotes.

Name: For the hyperbola $\frac{(x+2)^{2}}{4}-\frac{(y-5)^{2}}{16}=1$ find the center, foci, endpoints of the transverse axis, and the equations of the asymptotes.
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The Answer of For the hyperbola $\frac{(x+2)^{2}}{4}-\frac{(y-5)^{2}}{16}=1$
find the center, foci, endpoints of the transverse axis, and the equations of the asymptotes.

For the Hyperbola (x+2)24−(y−5)216=1\frac{(x+2)^{2}}{4}-\frac{(y-5)^{2}}{16}=14(x+2)2​−16(y−5)2​=1 Find the Center, Foci, Endpoints of the Transverse Axis, and the Transverse

For the Hyperbola $\frac{(x+2)^{2}}{4}-\frac{(y-5)^{2}}{16}=1$ Find the Center, Foci, Endpoints of the Transverse Axis, and the Transverse