Chapter 10 - Parametric Equations and Polar Coordinates - 10.5 Exercises - Page 700: 29

Hyperbola Foci: $( 0,\pm \sqrt 5-1)$ Vertices: $(0,\pm2-1)$

$y^{2}+2y=4x^{2}+3$ $\frac{(y+1)^{2}}{2^2}-\frac{x^{2}}{1^2}=1$, the equation of a hyperbola. $c^{2}=4+1$ $c=\sqrt 5$ Foci: $( 0,\pm \sqrt 5)$ Vertices is: $(0,\pm2)$ If we shift the hyperbola by one unit downwards, then: Foci: $( 0,\pm \sqrt 5-1)$ Vertices is: $(0,\pm2-1)$