Finding the eccentricity of the hyperbola $\left|\sqrt{(x-3)^2+(y-2)^2}\right|-\sqrt{(x+1)^2+(y+1)^2}=1$

Question

Find the eccentricity of the given hyperbola: $$\left|\sqrt{(x-3)^2+(y-2)^2}\right|-\sqrt{(x+1)^2+(y+1)^2}=1$$

(original problem image)

Answer 1

First off, the absolute value does not matter, because square roots are always positive. Because of the distance formula, this means: Distance from $(x,y)$ to $(3,2)$ minus distance from $(x,y)$ to $(-1,-1)$. Therefore, by the definition of a hyperbola, the foci are $(3,2),(-1,-1)$, and the distance from the center to the vertices is $\frac{1}{2}$.

This center is at the midpoints of the 2 foci, or $(1,0.5)$. The distance from a focus to the center is 2.5, by the distance formula.

Therefore, the eccentricity is $\frac{\mathrm{distance}\:\mathrm{from}\:\mathrm{focus}\:\mathrm{to}\:\mathrm{center}}{\mathrm{\mathrm{distance}\:\mathrm{from}\:\mathrm{vertex}\:\mathrm{to}\:\mathrm{center}}}=\frac{\frac{5}{2}}{\frac{1}{2}} = 5$

Hopefully this helps!